Math VI (English).pdf

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Mathematics 6
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In charge in-chief
Qadir Bux Rind
Chairman Sindh Textbook Board
Written by
Provincial Review Committee (PRC)
Mr. Arjan Lal S. Sudheria
Mr. Muhammad Sagheer Shaikh
Mr. Aftab Ali
Mr. Haroon Laghari
Ms. Attia Tabassum Bhutto
Syed Afaq Ahmed
Prof. Aijaz Ali Subehpoto
Mr. Nazir Ahmed Shaikh

C O N T E N T S
C O N T E N T S
C O N T E N T S
C O N T E N T S
C O N T E N T S
C O N T E N T S
1
WHOLE NUMBERS
SETS
FACTORS AND MULTIPLES
INTEGERS
SIMPLIFICATIONS
RATIO AND PROPORTION
INTRODUCTION TO ALGEBRA
FINANCIAL ARITHMETIC
LINEAR EQUATIONS
GEOMETRY
PERIMETER AND AREA
THREE DIMENSIONAL SOLIDS
Units Description Page No.
7
8
10
11
12
9
6
5
4
3
13 INFORMATION HANDLING
14 GLOSSARY
15 ANSWERS
2
1
16
36
64
88
102
119
137
154
166
194
214
228
243
247

P R E F A C E
P R E F A C E
P R E F A C E
P R E F A C E
P R E F A C E
P R E F A C E
The Sindh Textbook Board is an organization charged
with the preparation and publication of textbooks in the
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Chairman
Sindh Textbook Board

1.1 SET
In mathematics collection of things
is called set.
Unit
1 SETS
In 19th century, George Cantor was
the first mathematician who gave the
concept of set.
George Cantor
Maths-6
1
Define set
In mathematics we describe a set as:
A set is a collection of ‘well defined’ and ‘distinct’ objects.
Whereas the word ‘Distinct’ means different objects.
The term ‘Well defined’ means a set must have some specific
property so, that it can easily be identified whether or not an
object belongs to the given set.
We often use some words in our
daily life that represent a
collection of things like dinner set,
sofa set, tea set, group of boys,
a team of players, a bunch of keys,
crowd of people, a cluster of
trees, a flock of sheep. The words
set, group, team, bunch, crowd,
cluster, flock are used to denote
collection of things in daily life.

Let us discuss some examples of collection.
(1)
(2)
(3)
The students of class VI of your school. It is well defined
because students of class VI of your school can only belong
to it. Each student has unique identity. Thus it is a set.
The collection of favourite books. It is not well defined
because a book may be favourite for one person but may
not be for another. Thus this collection is not a set.
Name of days of week; Monday, Tuesday, Wednesday,
Thursday, Friday, Saturday, Sunday. It is a well defined
sets because every day is the member of this set, which
is distinct.
Teacher’s Note
Teacher should give some more examples of collections from class
room and daily life to clarify the concept of set such as name of books
in a bag for grade VI and name of instruments in geometry box..
1
2 Maths-6
SETS
Unit 1
English alphabets A, B, C, . . ., and Z are used to denote sets.
The objects of the set are called elements or members of the set.
Recognize notation of a set and its objects/elements
The symbol ‘Î’ is used to show the membership to a set and read
as “belongs to” and ‘Ï’ is used for, “does not belong” to a set.
Î
Ï
The elements of the set are listed within brackets { } and separated
by commas.
Let us consider the following examples.
Example 1:
(i) B = { 2, 4, 6, 8 }
The members of set B are 2, 4, 6 and 8.
As 2 belongs to set B
So, symbolically we write as 2 Î B
As 3 does not belong to Set B.
So, symbolically we write as 3 Ï B

SETS
Maths-6
Unit 1
3
(ii) D = { x, y, z }
The elements of set D are x, y and z.
As we have already described that all the elements of a set are
listed within { } and separated by commas, this form of describing
a set is called tabular form.
Example 1: Set of the vowels of the English alphabet
In tabular form we write as {a, e, i, o, u} or {i, a, u, e, o}
Example 2: Set of natural numbers upto 100
In tabular form we write as {1, 2, 3, ..., 100}
Example 3: Set of names of three boys whose names start
with “A”.
In tabular form we write as {Ali, Ahmed, Ahmer}
Here, x Î D, but p Ï D
Example 4: Set of name of days of a week.
In tabular form we write as
Example 5: Explain, why are the following not sets:
A = {5, 6, 6, 7, 8}
B = Collection of intelligent students
A = {5, 6, 6, 7, 8}
This is not set because the element 6 is repeated.
B = Collection of intelligent students
This is not set because it can not be decided that which
students are intelligent.
(1)
(2)
(1)
(2)
Solution:
Describe tabular form of a set and demonstrate through
examples
Let us demonstrate this form through examples.
{Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}

SETS
Unit 1
Points to remember
The name of a set is denoted by capital letter of
English alphabet.
Every element of a set is separated by a comma ‘,’ in
tabular form.
All the members of set are enclosed in brackets { } in
tabular form.
The order of writing elements does not matter in a set.
Elements of set do not repeat itself.
Write the given sets in tabular form.
Set Tabular Form
(i)
(ii)
(iii)
{ a, b, c, d, e }
Set of first five English alphabets.
Set of all factors of 12
Set of names of days of a week
which start from “F”
Teacher’s Note
Teacher should perform the given activities on blackboard with
students and also give some more examples.
1
4 Maths-6
(iv) Set of first five natural numbers
(i) (iii)
(ii) 6 ______ A
4 ______ A
Î 1 ______ A
(iv) 9 ______ A (v) 2 ______ A (vi) 8 ______ A
Activity 1
Activity 2
For A = {2, 4, 6, 8, 10 }
Fill in the blanks with symbols Î or Ï.

SETS
Maths-6
Unit 1
1.
3.
4.
2.
Tick ( ) which of the following are sets?
Write in tabular form.
If P = {a, e, i, o, u} and Q = {1, 2, 3, ..., 10} then which
of the following statements are true or false.
If A = {5, 6, 7, 8} and B = {a, b, c, d}, fill in the blanks
with the symbols Î or Ï.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(i)
(iv)
(i)
(iii)
(v)
(vii)
(ii)
(v)
(ii)
(iv)
(vi)
(viii)
(iii)
(vi)
Collection of the names of Presidents of Pakistan.
Collection of names of captains of hockey teams of Pakistan.
Collection of delicious dishes.
Collection of intelligent students in your class.
Collection of greater numbers.
Collection of English teachers in your school.
Set of name of capitals of the provinces of Pakistan.
Set of natural numbers from 50 to 70.
Set of first ten natural numbers exactly divisible by 2.
Set of letters in the word “pakistan”.
Set of mathematics teachers in your school.
Set of name of colours of national flag.
5 ______ A
c ______ B
a Î P ________
i Î Q ________
y Î Q ________
3 Î P ________
9 ______ A
b ______ B
1 Î Q ________
v Î P ________
e Î P ________
7 Î Q ________
m ______ A
7 ______ B
EXERCISE 1.1
5

Activity Tick ( ) the finite sets and cross ( )
û
ü
the infinite sets.
SETS
A set which has limited number of elements is called finite set.
Consider the sets given in the table 1.
Let us consider set A. It has limited number of elements, so it
is finite set.
A = {2, 4, 6} , B = {1, 2, 3, ...}
C = {6, 4, 2} , D = {x, y, z}
E = {1, 3, 5, 7, ...} , F = {5, 10, 15, 20, 25, ...}
A set which has unlimited number of elements is called
infinite set.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
Set of all citizen of Pakistan
Set of all whole number less than 9999
Set of drops of water in river
The set of all English alphabets
{1, 2, 3, ...}
{d, o, r}
{2, 4, 6, ...}
Define finite and infinite sets
Table 1
Maths-6
Unit 1
In table 1, C and D are also finite sets.
Now consider set B = {1, 2, 3, ...}.
This set has unlimited number of elements. So it is infinite set.
In table 1, E and F are also infinite sets.
6

SETS
Maths-6
Unit 1
P = { 8 } Q = { a }, R = {1} and S = {0}
Examples:
A set having only one element is called singleton.
In table 2, D is also empty set. Some other examples of
empty set are:
Table 2
Examples:
7
A set which has no element is called an empty or null or
void set. It is denoted by or { }.
(i) Set of squares with five sides.
(ii) Set of letters of English alphabet before A.
In table 2, there is a set A = { 5 } which has only one element
and is called singleton set.
Define empty/void/null sets and singleton
A = { 5 } , B = { x }
C = Set of natural numbers less than 1
D = Set of triangles with four sides
In table 2, C has no element because there is no natural number
less than 1.
So, C is an empty set, which is also called null or void set.
In table 2, set B is also singleton. Some other examples of
singleton are:

SETS
Unit 1
Also, in the given table 3,
A ~ C and B ~ F
For example in table 3, (i) A = C and also A ~ C
(ii) D ~ E but D ¹ E.
All equal sets are also equivalent sets but all equivalent
sets may not be equal sets.
Table 3
Consider sets A and C from the table 3. All the elements
of both sets are the same. So they are called equal sets.
A = {1, 2, 3, 4} , B = {x, y, z}
C = {4, 3, 2, 1}, D = {a, e, i, o, u}
E = {2, 4, 6, 8, 10}, F = {y, z, x}
Define equal and equivalent sets
In table 3, B and F are also equal sets.
When we see D and E, they have same number of elements and
these are called equivalent sets.
Two sets A and B are called equal sets if they have all elements
same. Symbolically we represent as A = B and read as:
Set A is equal to set B
Two sets A and B are called equivalent sets if the number
of elements of both sets are same. Symbolically we write as:
~
A B and read as: Set A is equivalent sets to set B
Teacher’s Note
Teacher should encourage students to make examples of different
types of sets themselves.
1
8 Maths-6

SETS
Unit 1
In the given table 4,
B Ê A and D Ê B
If A is subset of B then the set B is called superset of A.
Symbolically we write it as ‘ B Ê A’
Table 4
Consider the sets given in the following table 4.
Let us see sets B and C. All the elements of B are also elements
of C. So we say B is subset of C.
In the given table 4,
A Í B, B Í C , C Í D and D Í C
As B Í B and D Í D
Again, if we consider B and D.
B is subset of D. So D is called superset of B.
Remember that
A = { 1 } , B = {1, 2, 3}
C = {1, 2, 3, 4}, D = {4, 3, 2, 1}
If each element of set A is also an element of set B then set A
is called subset of set B. Symbolically we write as A Í B.
Every set is subset of itself.
Define subset and superset of a set
Maths-6
9

SETS
Maths-6
Unit 1
10
Define proper and improper subsets of a set and demonstrate
through examples.
Identify and write the following sets,
from the given table.
(i)
(iii)
(v)
(vii)
(ii)
(iv)
(vi)
(viii)
Two finite sets
Two empty sets
Two equivalent sets
Two subsets
Two infinite sets
Two equal sets
Two singleton
Two super sets
Set of natural numbers between 2 and 3, Set of factors of 1
Set of triangles with four sides, D {2, 4, 6, 8}
=
A {1, 3, 5, 7, ...}, B { 6 }, E {5, 10, 15, 20, 25, ...}
= =
=
C {a, b, c, d}, F {4, 6, 2, 8},
= =
Set of first four letters of English alphabet
Proper Subset
Improper Subset
In the given table 4, A Ì B and B Ì D
In table 4, we see that set D is subset of set C but set D is also
equal to set C. So D is called improper subset of C.
In table 4, set C is improper subset of set D.
If set A is subset of set B and also A is not equal to B then
“A is proper subset of set B. Symbolically we write as A Ì B.
If set A is subset of set B and A is also equal to B then “A is
called improper subset of set B”.
Activity
Now we consider set B and set D in table 4
Here B Í D and B ¹ D.
So set B is called proper subset of set D.

SETS
Maths-6
Unit 1
11
EXERCISE 1.2
3.
4.
Which of the following are empty sets?
Which pairs of the following are equal sets?
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(i)
(iii)
(iv)
(v)
(ii)
2. Which of the following are finite or infinite sets?
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
A { 0, 1, 2, ..., 50 }.
=
B { 100, 200, 300, 400, ... }.
=
Set of hair on the body of goat.
Set of legs of a cat.
Set of stars in the sky.
Set of lines passing through a point.
Set of natural numbers greater than 20.
Set of all the cities of Pakistan.
Set of all the schools in Sindh.
Set of all even numbers.
1. Give reasons why following collections are not set.
(i) (ii)
(iv)
{ d, o, o, r }
(iii) { a, f, d, a }
{ «, O, «, D }
{ 2, 2, 3, 3, 4, 4 }
Set of students of your class over 20 years of age.
Set of letters after Z in English alphabet.
Set of children whose names start with “K” in
your locality.
Set of name of solar calendar starting with Z.
Set of fishes which live in sand.
Set of even numbers between 4 and 10.
{ 0 } (viii) { }
{ 1, 2, 3 } and { 2, 3, 1 }
{ k, i, t, e } and { b, i, t, e }
{ x, y, z } and set of first three letters of English alphabet.
Set of odd numbers less than 2 and { }
{ p, a, t } and { t, a, p }
(v) Set of beautiful birds (vi) Set of good players

SETS
Maths-6
Unit 1
12
6.
7.
Which of the following statements are true or false?
Choose appropriate symbol to indicate the relation
between sets ( , Í , Ë , Ê , ~,Ì)
=
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
B = { m, o, n } if B is Set of letters of word “MOON”.
(i)
A Í B if A = { 1, 2 } and B is { 2, 3, 4, 5 }
(ii)
If A = { 4, 5, 10 } and B is { 5, 10, 20 } then A ~ B.
(iii)
If A = { 0, 1, 2, 3, 5 } and B is Set of natural numbers
less than 6, then A = B.
(iv)
If X is equal to Y, then X is equivalent to Y.
(v)
{ 10, 20, 30 } ______ { 5, 10, 15, 20, 25, 30 }
{ 7, 14, 21, 28, 35, 42, 49 } _______ Set of
multiples of 7 less than 50.
{ 5, 6, 7, 8, ... } _______ { 7, 5, 8 }
{ 0 } _______ { }
{ t, e, a } _______ { a, e, t }
{ Hyderabad, Karachi, Sukkur }_______Set of
all cities of Sindh.
{ a, e, i, o, u } _______ Set of letters of English
.
alphabets
{ 11, 22, 33 } _______ { 1, 2, 3, ... }
5. Which pairs of the following are equivalent sets?
(i)
(iii)
(iv)
(ii)
{ m, i, l, e } and { n, i, l, e }
{ D, , , «, ÿ } and { 1, 2, 3, 4, 5 }
{ 6, 66, 666, 6666 } and { 666, 7777, 77 }
{ 1, 2, 3 } and {3, 2, 1}

SETS
Maths-6
Unit 1
13
3. Write two examples of set.
REVIEW EXERCISE 1
4. Write the following sets in tabular form.
A is Set of all even numbers between 2 and 10.
B is Set of all odd numbers from 1 to 17.
C is Set of natural numbers divisible by 2 and less than 30.
D is Set of months of a solar calendar beginning with letter “J”.
(i)
(ii)
(iii)
(iv)
1.
2.
Fill in the blanks.
Choose the correct answer.
If A = { a, b, c } then a, b, c are __________ of a set A.
If B is set of vowels of English alphabet:
{ f } is called:
If sets A and B are equal, we use symbol
If A = { a, b, c } and B = { a, b, c, d, e } then
(a)
(a)
(a)
(a)
(b)
(b)
(b)
(b)
(c)
(c)
(c)
(c)
(d)
(d)
(d)
(d)
p Î B
null set
Î
A = B
e Ï B
infinite set
Ì
A ~ B
b Î B
subset
~
A Ì B
a Î B
singleton set
=
B Ì A
A set which has no element is called _____________ set.
A set which has limited number of elements is called
a ______ set.
If two sets A and B have ______________ elements, then
they are said to be equal sets.
(i)
(i)
(ii)
(iii)
(iv)
(ii)
(iii)
(iv)

SETS
Maths-6
Unit 1
14
7. If A = { d } , B = { c, d } , C = { a, b, c } , D = { a, b } then
which of the following statements are true or false.
D Í C A Ë C B Í D
(i) (ii) (Iii) C Ë D
(iv)
5. Which of these sets are finite and which of these are
infinite.
A is Set of members of your family
B is Set of all even numbers
C is Set of prime factors of 60
D is Set of all multiples of 7
E is Set of one-digit natural numbers
(i)
(ii)
(iii)
(iv)
(v)
6. Which of the following are empty sets.
L is Set of the names of days of a week which start with “S”.
F is Name of months of a year having 32 days.
M is Set of odd numbers exactly divisible by 2.
N is Set of prime numbers exactly divisible by 2.
(i)
(ii)
(iii)
(iv)
8. Indicate which of the following statements are true and
false?
The members of any set are always of same kind.
The order of the elements of a set does not matter.
An object can be included in a set repeatedly.
Any collection of objects is called a set.
The notation “Î” is a symbol to indicate membership
of a set.
A well-defined collection of distinct objects is called
a set.
{ a, b, c, d } is not a set.
The notation “Ï” is a symbol to indicate membership
of a set.
1, 2, 3, 4 is a set.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)

SETS
Maths-6
Unit 1
15
A set is a collection of well defined and distinct objects.
Each object in a set is an element or member of the set.
A set is described in two ways.
(i) Precisely described in words.
(ii) All the elements are listed down separately.
Finite set has limited number of elements.
Infinite set has unlimited number of elements.
A set having no element is an Empty or Null set.
Two sets are equal if they have all same elements.
Two sets are equivalent if they have same number of elements.
B is subset of A if every element of B is also an element of A.
A set having only one element is called singleton set.
SUMMARY
Symbols to Remember
is member of
is not member of
or { } empty set or null set
is equal to
is not equal to
is equivalent to
is a subset of
is a proper subset of
is a superset of
is not a subset of
Î
Ï
=
~
Í
Ì
Ê
=
Ë

1
16 Maths-6
2.1 NATURAL AND WHOLE NUMBERS
We use numbers ‘1, 2, 3, 4, 5, ...’ for representing the quantity
of objects in our daily life.
These numbers 1, 2, 3, 4, ..., are called natural numbers.
We denote the set of natural numbers by N.
i.e N = { 1, 2, 3, 4, 5, 6, ...}
Look at the following sets:
N = {1, 2, 3, ...}
W = {0, 1, 2, ...}
Important facts about natural numbers and whole numbers are:
Identify natural and whole numbers, and their notations
When we include ‘0’ in the set of natural numbers, we will get
the set of whole numbers.
‘0, 1, 2, 3, 4, ...’ are called whole numbers.
Here 0 is predecessor of 1 and 1 is successor of 0. Similarly 1
is predecessor of 2 and 2 is successor of 1 and so on.
Unit
2 WHOLE NUMBERS
The set of whole numbers consist of zero and all natural
numbers. It is denoted by W,
i.e. W = { 0, 1, 2, 3, 4, 5, 6, 7, ... }
1 is the smallest and the first natural number.
The set of natural numbers is an infinite as we can not
count all natural numbers.
The first and the smallest whole number is ‘0’.
The set of whole numbers is an infinite set.
The only difference between set of natural numbers and set of
whole numbers is of number ‘0’, ‘0’ does not belong to the set of
Natural numbers. All the other members of both the sets are the
same.
Differentiate between natural and whole numbers

Number line helps us to represent given whole numbers.
Represent a given list of whole numbers
1
17 Maths-6
Whole numbers can be represented on a number line by using
following steps:
Note:
(i)
(ii)
On a number line, a number is greater than any number
on its left. For example, 1 > 0, 2 > 1 and 3 > 2 etc.
On a number line a number is less than any number on
its right. For example 0 < 2, 1 < 3, 5 < 12.
(ii)
(i)
Steps:
Let us represent numbers 0, 1, 2, 3, 4, 5, ..., on a number line.
(iii)
Thus we get a number line as under:
Mark a point to represent the first whole number ‘0’ on the line.
Draw a line
Mark other points at equal distance and name them 1, 2,
3, 4, 5, ... respectively as shown below:
0 1 2 3 4 5 6 7 8 9 10 11 12
0
Example.
Represent the following on number line.
Let us represent whole numbers < (or >) a given whole number on
number line with the help of an example.
Solution:
Number less than 6 means number to the left side of 6.
(i)
(i) Whole numbers less than 6
Whole numbers Less than 6
(ii) Whole numbers Greater than 4
0 1 2 3 4 5 6 7
Represent whole numbers < (or >) a given whole number on a
number line
Unit 2 WHOLE NUMBERS

Teacher’s Note
Teacher should give more questions to students on blackboard to
represent whole numbers on number line.
1
18 Maths-6
Dark shaded dots represent the required whole numbers.
So, 0, 1, 2, 3, 4, 5 are the whole numbers less than 6.
Dark shaded dots represent the required whole numbers.
So, 5, 6, 7, ... are the required whole numbers.
Dark shaded points represent the required whole numbers.
(ii) Whole numbers greater than 4
0 1 2 3 4 5 6 7
Let us represent whole number > (or <) a given whole number on
a number line with the help of an example.
Solution: (i) whole number £ 7
So, the whole numbers less than or equal to 7 are 0, 1, 2, 3, 4,
5, 6, 7.
(i)
(ii)
Whole numbers less than or equal to 7
Whole numbers greater than or equal to 3
0 1 2 3 4 5 6 7 8 9 10
Represent whole numbers > ( or < ) a given whole number on
a number line
(ii) Whole number ³ 3
Dark shaded points represent the required whole numbers with
continuation.
So, the whole numbers greater than or equal to 3 are 3, 4, 5, 6, 7, ...
0 1 2 3 4 5 6 7
Example. Represent the following whole number on a number line.

Example. Represent the even whole numbers greater than 1 but
less than 13 on number line.
Dark shaded points represent the required even whole numbers.
Dark shaded points represent the required odd whole numbers.
Solution: Even whole numbers > 1 but < 13
Hence the required even numbers are 2, 4, 6, 8, 10, 12.
0 1 2 3 4 5 6 7 8 9 10 11 12 13
Solution: Odd whole numbers ³ 5 but £ 11
Example: Represent odd whole numbers greater than or equal to
5 but less than or equal to 11 on a number line.
Hence the required odd whole numbers are 5, 7, 9, 11.
0 1 2 3 4 5 6 7 8 9 10 11 12 13
Teacher’s Note
Teacher should draw the number line on blackboard and move a
student in front of it to show the sum of numbers on number line.
1
19 Maths-6
Represent whole numbers > but < a given whole number on
a number line
Let us represent whole numbers > but < a given whole number
on a number line with the help of an example.
Represent whole numbers > but < a given whole number on
a number line
Represent sum of two or more given whole numbers on the
number line
The method of adding two or more given whole numbers on the
number line is explained with the help of following example.
Let us represent whole numbers > but < a given whole number on
a number line with the help of an example.

1
20 Maths-6
Example 2.
With the help of a number line, find the sum of 1 + 3 + 5
Start from 0, and move 6 units on the right. Again, starting from
6 move 2 units on the right. We reach at number 8.
Hence, 6 + 2 = 8
Therefore, 1 + 3 + 5 = 9
Solution:
Its procedure is described below:
0 1 2
1 + 3 + 5
3 4 5 6 7 8
9
9 10 11
1. Write, if possible:
Write first ten whole numbers.
Write first ten natural numbers.
3.
2.
(i)
(ii)
(iii)
(iv)
The smallest natural number
The smallest whole number
The largest natural number
The largest whole number
EXERCISE 2.1
Example 1. Find the sum of 6 and 2 by using number line.
Solution:
0 1 2 3 4 5 6 7
+ 2
6
8
8
9 10

1
21 Maths-6
4. Represent on number line:
(i)
(v)
(ii)
(iii) (iv)
Whole numbers > 8
Whole numbers > 3 but < 15
Whole numbers < 8
(vi) Whole numbers > 5 but £ 12
Whole numbers ³ 8
(vii)
(ix)
(x)
Whole numbers ³ 4 but £ 11
Odd whole numbers greater than 3
Even whole numbers greater than or equal to 8 but
less than 16.
(viii) Whole numbers ³ 3 but < 9
Whole numbers £ 8
5. Find sum of the following whole numbers by using
number line.
(i)
(v)
(iii)
(ii)
(vi)
(iv)
1 and 5
2, 3 and 5
10 and 2
6 and 3
3, 2 and 4
8 and 4
2.2 ADDITION AND SUBTRACTION OF
WHOLE NUMBERS
How to fill the triangle with
1, 2, 3, 4, 5, 6, 7, 8 and 9 to
get the sum of each side equal
to 17.
We already know how to add
and subtract two natural
numbers.
Add and subtract two
given whole numbers
Same methods are applied for
whole numbers.
For example, the sum of 5 and 14 is 19 which is again a whole
number. Therefore we can say that the sum of whole numbers is
always a whole number.

1
22 Maths-6
When we subtract 23 from 53, we get 30. Here 23, 53 and 30 are
all whole numbers. Hence, the resulting number in subtraction
of whole numbers may or may not be a whole number.
Similarly, 25941 + 58723 = 84664 which is a whole number.
The sum of two whole numbers is always a whole number.
Example 1: Add: 389 and 245
Solution: Adding the monthly income of Aslam and his wife.
Monthly income of Aslam
Monthly income of his wife
Total of monthly income
Monthly income
Monthly expenditure
Monthly saving
Now, in order to find monthly saving, we subtract monthly
expenditure from their total income.
Hence, their monthly income is Rs 92435 and monthly
saving is Rs 27435.
=
=
=
=
=
=
56835 rupees
+ 35600 rupees
92435 rupees
92435 rupees
65000 rupees
27435 rupees
Example 3: Aslam earns Rs 56835 in a month and his wife earns
Rs 35600. Their monthly expenditure is Rs 65000. Find their
total monthly income and saving.
Hence 389 + 245 = 634
3 8 9
+ 2 4 5
6 3 4
1
1
As,
9 + 5 = 14
1 + 8 + 4 = 13
1 + 3 + 2 = 6
Hence 1000 – 535 = 465
1 0 0 0
– 5 3 5
0 4 6 5
9 1
9
0
As,
10 – 5 = 5
9 – 3 = 6
9 – 5 = 4
0 – 0 = 0
Example 2: Subtract 535
from 1000

1
23 Maths-6
Example 4: Find the sum of smallest five digit number and largest
four digit number.
Smallest five digit number
Largest four digit number
Hence, required sum = 19999
1 0 0 0 0
+ 9 9 9 9
1 9 9 9 9
Solution:
The sum of two whole numbers, in any order, is always same.
Verify commutative and associative law (under addition)
of whole numbers.
It shows that, for any order of two given whole numbers, their
addition give us the same sum. This is known as the commutative
law of whole numbers under addition.
When we add 9 to 30, we get 39 and when we add 30 to 9, then
again we get 39. Similarly by adding 100 to 40 and by adding
40 to 100, we get the same whole number 140. In other words,
we have
Commutative law under addition.
(a)
9 + 30 = 30 + 9, and 100 + 40 = 40 + 100
Thus commutative law under addition is:
Example:
Verify commutative law under addition for whole numbers 85 and 95.
Solution: By Commutative Law under addition, 85 + 95 = 95 + 85
LHS = 85 + 95
= 180
Since, LHS = RHS, therefore 85 + 95 = 95 + 85
Hence, Commutative law under addition is verified.
RHS = 95 + 85
= 180
8 5
+ 9 5
1 8 0
9 5
+ 8 5
1 8 0

Teacher’s Note
Teacher should help the students to understand the concepts of
laws of whole numbers under addition.
1
24 Maths-6
Hence, three whole numbers added in any order give same result.
We can add only two whole numbers at a time. So, we can get
the sum of three whole numbers, say 3, 7 and 12 by two ways:
(b) Associative law under addition.
This is known as associative law under addition.
Thus Associative law under addition is:
The sum of three whole numbers, in any order, is always same.
Therefore we can say
(3 + 7) + 12 = 3 + (7 + 12)
or
3 + 7 + 12
= 10 + 12
= 22
3 + 7 + 12
= 3 + 19
= 22
Example:
Verify Associative law under addition for 23, 59 and 87.
Solution:
By Associative Law under addition, (23 + 59) + 87 = 23 + (59 + 87)
LHS = (23 + 59) + 87
= 82 + 87
= 169
Since, LHS = RHS, therefore (23 + 59) + 87 = 23 + (59 + 87)
So, Associative law under addition is verified.
RHS = 23 + (59 + 87)
= 23 + 146
= 169
2 3
+ 5 9
8 2
+ 8 7
1 6 9
5 9
+ 8 7
1 4 6
+ 2 3
1 6 9

1
25 Maths-6
(i)
(ii)
(iii)
(iv)
56123
56123 + 71045 = 71045 + __________
24125 + (41625 + 7123) = ( 24125 + _________ ) + 7123
47813 + ___________ = 51623 + ___________
567 + ( __________ + 1784) = ( __________ + 962) + _________
Activity 1 Fill in the following blanks by using
Commutative and Associative laws of addition.
Recognize ‘0’ as additive identity
For Example: 1 + 0 = 0 + 1 = 1
25 + 0 = 0 + 25 = 25
And the number “0” is called additive identity.
There is a whole number, which has a unique property that no
other whole numbers has. The number is 0, and the property is:
When 0 is added to any whole number, the sum is the whole
number itself.
3.
4.
Arif deposited Rs 45800 in his bank account. After a
month he withdrew Rs 3500 from it. How much money
was left in his account?
Kashif had Rs 82000. He gave Rs 6500 to his wife,
Rs 10550 to son and Rs 15335 to daughter. How much
money was left with him?
2.
1.
Population of a village is 2700. If 1070 are men and 915
are women, find the number of children.
What is the sum of the largest number of six digits and
the smallest number of seven digits?
EXERCISE 2.2

6.
5.
Verify associative law under addition for the following.
Verify commutative law under addition for the following.
(i)
(iii)
(ii)
(iv)
(i)
(iii)
(ii)
(iv)
34006, 2389 and 44380
231, 6090 and 25996
583031, 127 and 3405
412, 3007 and 102341
7628 and 39780
29000 and 10699
924981 and 228
50102 and 9019854
7.
8.
Fill in the blanks to make the following statements true.
Think a number. Double the number and add 9 to it then
add the number you started with. Then divide the amount
by 3 and also subtract 3 from the quotient. What the
number do you get?
(i)
(ii)
(iii)
(iv)
(v)
5020 + 849 = ________________ + 5020
97864 + 0 = ________________
749 + 0 = 0 + ______ = ______
3971 + (430 + 300) = (3971 + ____________) + 300
4853 + 93 = 93 + 4853 = _________________
2.3 MULTIPLICATION AND DIVISION OF
WHOLE NUMBERS
We know that multiplication is repeated addition and division is
repeated subtraction.
Multiply and divide two given whole numbers
(a) Multiplication:
The method of multiplying two whole numbers is same as the
method of multiplication of natural numbers.
1
26 Maths-6

1
27 Maths-6
In all these examples, the multiplication of two whole numbers
gives a whole number.
5 ´ 4 = 20, 10 ´ 153 = 1530, 0 ´ 2453 = 0
Consider the following multiplications:
The product of two whole numbers is always a whole number.
Let us consider the following examples.
Example 1:
Multiply 2483 by 253
Example 2: Zahida purchased 20
bangles at the price of Rs 80 each.
Find the total amount paid by
Zahida.
Solution:
Solution:
Hence the total amount is Rs 1600.
Hence
2483 ´ 253 = 628199
Number of bangles:
Price:
2 0
´ 8 0
0 0
1 6 0 ´
2 4 8 3
´ 2 5 3
7 4 4 9
1 2 4 1 5 0
4 9 6 6 0 0
6 2 8 1 9 9
1 6 0 0
Let us consider division of some other whole numbers
along with their quotients and remainders.
Dividend
9
7
14
56
3
5
8
7
3
1
1
8
0
2
6
0
Divisor Quotient Remainder
If we divide 9 by 2, then the quotient is 4 and the
remainder is 1.
(b) Division:
The quotient of a whole number divided by 1 is the
number itself.
Note:
The quotient of zero divided by a non-zero whole number
is zero.
9
– 8
1
4
2

1
28 Maths-6
Example 1: Divide 435 by 3. Write quotient and remainder.
Solution:
Hence, 435 ¸ 3 = 145.
15
– 15
0
435
– 3
13
– 12
145
3
divisor
quotient
dividend
remainder
Example 2: The cost of 15 books is Rs 600. Find the cost of one book.
Solution: In order to find the cost of one book we divide the
total cost Rs 600 by number of books, i.e. 15.
So, the cost of one book is Rs 40.
600
– 600
´
40
15
Example 3:
Find the greatest 4 digit number which is exactly divisible by 23.
Solution:
As greatest 4 digit number is 9999
Thus the required number = 9999 – 17
= 9982
109
– 92
17 = remainder
9999
– 92
79
– 69
434
23

1
29 Maths-6
1. Find the following products of whole numbers.
(i)
(iii)
(ii)
(iv)
854 ´ 96
256 ´ 1008
736 ´ 103
995 ´ 158
EXERCISE 2.3
6.
4.
5.
3.
Multiply the greatest number of four digits with the
smallest number of three digits.
The shopkeeper purchased 125 television sets. If the cost
of each set is Rs 9820, find the cost of all sets.
Find the greatest 5 digit number which is exactly divisible
by 75.
A gardener plans to plant 570 trees in 19 rows. Each row
should contain equal number of trees. How many trees
will be in each row?
2. Divide and find the quotient and remainder.
(i)
(iii)
(ii)
(iv)
7772 ¸ 58
16025 ¸ 1000
96324 ¸ 245
92845 ¸ 300
The product of two whole numbers, in any order, is always same.
Verify commutative and associative law
(under multiplication) of whole numbers
(a) Commutative law under multiplication
Consider the product of two whole numbers 5 and 6, which is
5 ´ 6 = 30. Also 6 ´ 5 = 30.
Thus 5 multiplied by 6 is the same as 6 multiplied by 5.
Similarly, 401 ´ 98 = 98 ´ 401 gives 39298 = 39298,
3756 ´ 23 = 23 ´ 3756 gives 86388 = 86388, 101 ´ 0 = 0 ´ 101
gives 0 = 0 and 255 ´ 1 = 1 ´ 255 gives 255 = 255.
Hence, two whole numbers multiplied in any order gives equal
product. This is known as commutative law under multiplication.

So “1” is called multiplicative identity.
30 Maths-6
Consider the product of three whole numbers 3, 2 and 4. If we first
multiply 3 and 2 and multiply the resulting number by 4, we get
Associative law under multiplication
On the other hand, if we multiply
2 with 4 and multiply their product
with 3.
(3 ´ 2) ´ 4 = 6 ´ 4 = 24.
The product of three whole numbers, in any order, is always same.
What are the things
we can do without any
definite order in our
daily life?
Hence, three whole numbers multiplied in any order gives equal
product. This is the associative law under multiplication.
Thus associative law under multiplication is:
Note:
Recognize ‘1’ as multiplicative identity
The whole number 1 has a special role in multiplication just as
the number 0 has in the case of addition. As 3 ´ 1 = 1 ´ 3 = 3
and 1 ´ 5 = 5 ´ 1 = 5 etc. Therefore 1 is the only whole number
with the property that when 1 is multiplied with any whole
number in any order, the result is the same whole number.
Therefore when 0 is multiplied with any whole number
the result is 0.
The number 0 also has a special property in multiplication.
We know that 8 ´ 0 = 0, 0 ´ 200 = 0 and 283450 ´ 0 = 0.
we get 3 ´ (2 ´ 4) = 3 ´ 8 = 24
Thus, (3 ´ 2) ´ 4 = 3 ´ (2 ´ 4)

1
31 Maths-6
2.4 MULTIPLICATION AND ADDITION
(SUBTRACTION) OF WHOLE NUMBERS
For the three whole numbers 2, 4
and 7. First we consider 4 + 7 and
then 2 ´ (4 + 7) to get 2 ´ 11 = 22.
Also (2 ´ 4) + (2 ´ 7) = 8 + 14 = 22
Therefore 2 ´ (4 + 7) = (2 ´ 4) + (2 ´ 7)
This is called distributive law of multiplication over addition.
Verify distributive law of multiplication over addition
What are the things
in our daily life must
be done in a certain
order?
Let us demonstrate the distributive law of multiplication over
addition with the help of example 5 (10 + 8) = 5 ´ 10 + 5 ´ 8
Solution:
L.H.S
Hence distributive law of multiplication over addition is verified.
R.H.S
=
=
=
L.H.S = R.H.S
5 (10 + 8) = 5 ´ 10 + 5 ´ 8
=
=
=
5 (10 + 8)
5 (18)
90
5 ´ 10 + 5 ´ 8
50 + 40
90
For the above three whole numbers 2, 4 and 7,
Therefore, 2 ´ ( 7 – 4) = ( 2 ´ 7) – (2 ´ 4)
2 ´ (7 – 4) = 2 ´ 3 = 6 and (2 ´ 7) – (2 ´ 4) = 14 – 8 = 6
This is called distributive law of multiplication over subtraction.
Let us demonstrate of Distributive law of multiplication over
subtraction with help of example 5 (20 – 2) = 5 ´ 20 – 5 ´ 2
Verify distributive law of multiplication over
subtraction (with positive difference)

1
32 Maths-6
1.
2.
Fill in the blanks.
Write True or False
(i)
(i)
(ii)
(iii)
(iv)
_________
_________
_________
_________
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
5 ´ 0 = __________
545 ´ 248 = 545 + 248
67 – (125 – 12) = (67 – 125) – 12
64 ´ (245 – 10) = 64 ´ 245 – 64 ´ 10
95 ´ (25 ´ 14) = (95 ´ 25) + 14
6 ´ 1 = __________
2 ´ 5 = 5 ´ ______
3 ´ (1 ´ 4) = (3 ´ 1) ´ _____
5 ´ ( _____ ´ 8) = ( _____ ´ 7) ´ 8
541 ´ (645 + ______ ) = 541 ´ 645 + 541 ´ 964
345 ´ (650 – 125) = 345 ´ 650 ______ 345 ´ ______
EXERCISE 2.4
Solution:
L.H.S
Hence distributive law of multiplication under subtraction is
verified.
R.H.S
=
=
=
L.H.S = R.H.S
5 (20 – 2) = 5 ´ 20 – 5 ´ 2
=
=
=
5 (20 – 2)
5 (18)
90
5 ´ 20 – 5 ´ 2
100 – 10
90

1
33 Maths-6
4.
6.
5.
3.
Verify commutative law under addition for 530 and 750.
Verify distributive law of multiplication over addition for
300, 615 and 975.
Verify Associative Law under multiplication for 240, 425
and 35.
Name and verify the law used in each of the following.
(i)
(iii)
(ii)
(iv)
9 ´ 7 = 7 ´ 9
5 ´ (7 – 1) = 5 ´ 7 – 5 ´ 1
4 ´ (6 ´ 3) = (4 ´ 6) ´ 3
2 ´ (8 + 3) = 2 ´ 8 + 2 ´ 3
REVIEW EXERCISE 2
1. Fill in the blanks by using laws under multiplication of
whole numbers.
(i)
(ii)
(iii)
(iv)
2 ´ _______ = 3 ´ ______
112 ´ 528 = _______ ´ ______
5 ´ (3 ´ 8) = (5 ´ 3) ´ _______
(vi)
(vii)
(viii)
_______ ´ 7 = 7 ´ _______ = 0
9 ´ (2 + 16) = _______ ´ 2 + _______ ´ 16
(20 + 4) ´ 1 = _______ ´ 1 + 4 ´ _______
17 ´ (8 – 3) = 17 ´ _______ – _______ ´ 3
(12 – 6) ´ 32 = 12 ´ _______ – 6 ´ _______
(v)
2. Fill in the blanks.
(i) 5 ´ 4 = ´ 5
(ii) 15 ´ (10 ´ 6) = ( ´ 10)
(iii)
(iv)
10 ´ (15 + 6) = (10 ´ 15) + ( ´ 6)
10 ´ (100 + 60) = ( ´ 100) + (10 ´ )

1
34 Maths-6
3.
4.
Write True or False.
Write the predessor and successor of the following
numbers:
(i)
(i) (ii)
(iii)
(v)
(vi)
5 – 3 = 3 – 5
671 245
2 – (0 – 8) = (2 – 0) + 8
(ii) 3 ´ 1 + 7 = 3 ´ 8
4 ´ (35 ´ 2) = (4 ´ 35) ´ 2
24 – (50 – 6) = (24 – 50) – 6
(iv) 9 + (7 + 5) = (9 + 7) + 5
(vii)
(ix)
14 ¸ 0 = 14 (viii) 0 ¸ 125 = 0
18 ¸ 18 = 0 (x) 75 ¸ 75 = 1
5. Choose the correct answer:
(i)
(ii)
The smallest natural number is _______.
The predecessor of 1 in the set of whole numbers
is ________.
(a)
(a)
(b)
(b)
(c)
(c)
(d)
(d)
0
0
1
2
2
3
100
none
(iii)
(iv)
(v)
The smallest seven digit number is ________.
The greatest six digit number is ____________
The numbers divisible by 2 are called_______numbers.
(a)
(a)
(a)
(b)
(b)
(b)
(c)
(c)
(d)
(d)
1234567
876543
prime
9999999
999999
even
111111
odd
(c) (d)
1111111 1000000
100000
whole
6. Draw a number line to represent the following whole numbers.
(i)
(iv)
(ii)
(v)
(iii)
0, 1, 3, 9
Whole numbers > 5 but < 10
Whole numbers > 3
Whole numbers ³ 1 but £ 8
Whole numbers £ 8

1
35 Maths-6
7.
8.
10.
11.
9.
Find the sum of 4, 2, 3 and 5 on number line.
Find two whole numbers whose sum is:
Verify Associative laws of addition and multiplication
for 20, 30 and 60.
Verify distributive laws of multiplication over addition and
subtraction for 700, 500 and 100.
Verify commutative laws of addition and
multiplication for 190 and 330.
(i) (ii) (iii)
9 24 31
12.
13.
What is the cost of 640 photocopies at Rs 1.50 per copy?
Find the greatest number of 4 digits which is exactly
divisible by 44.
Natural numbers are used for counting.
Some of the set of numbers and their notations are:
Set of natural numbers, N = {1, 2, 3, ...}
Set of whole numbers, W = {0, 1, 2, ...}
SUMMARY
Sum and product of two whole numbers is also a whole
number.
Zero is additive identity in the set of whole numbers.
Addition of whole numbers is commutative and associative.
Multiplication of whole numbers is commutative and
associative.
The multiplicative identity in the set of whole numbers is 1.
Multiplication is distributive over addition and subtraction
(with positive difference) in the set of whole numbers.
Division of two whole numbers is a whole number only if
divisor completely divides the number and remainder
is zero.

3.1 FACTORS AND MULTIPLES
Define a factor as a number which divides the dividend
completely leaving no remainder.
We know that if a dividend is divided by a divisor, we get quotient
and remainder.
Let us consider division of 8 by 2 and 3 separately.
Here 2 is a factor of 8 because remainder is zero or 2 divides 8
exactly. 3 is not a factor of 8 because remainder is not zero
or 3 does not divides 8 exactly.
Unit
3
Division by 3
Division by 2
2 3
8 8
– 8 – 6
0 2
4 or 8 ¸ 2 = 4
remainder = 0
or 8 ¸ 3 = 2
remainder = 2
2
A factor is a number which divides the dividend
completely leaving no remainder.
Maths-6
36
(a) (b) (c) (d)
2 4 5 6
Example: Which of the following are the factors of 10.
Solutions:
So, 2 and 5 are the factors of 10.
(a)
10 = 2 ´ 5
Since Remainder = 0
So, 2 is a factor of 10
2 10
– 10
0
5
(c)
10 = 5 ´ 2
Since Remainder = 0
So, 5 is a factor of 10
5 10
– 10
0
2
Since Remainder is not zero
So, 4 is not a factor of 10
(b) 4 10
– 8
2
2
Since Remainder is not zero
So, 6 is not a factor of 10
(d) 6 10
– 6
4
1

(a) 12
(a)
So, 1, 2, 3, 4, 6 and 12 can
divide 12 exactly.
Hence 1, 2, 3, 4, 6 and 12 are
factors of 12.
12
12 = 1 ´ 12
12 = 2 ´ 6
12 = 3 ´ 4
(b)
So, 1, 2, 4, 5, 10 and 20 can
divide 20 exactly.
Hence 1, 2, 4, 5, 10 and 20 are
factors of 20.
20
20 = 1 ´ 20
20 = 2 ´ 10
20 = 4 ´ 5
As
As
(b) 20
Example 2: Find all the factors of:
Solution:
Unit 3
The number itself and 1 are always the factors of the given number.
Teacher’s Note
Teacher should help the students to differentiate between factors
and multiples by giving more examples.
1
37 Maths-6
So, a number has always unlimited number of multiples.
Define a multiple as a dividend into which a factor
can divide
Let us consider the division of 18 by 2.
Here the divisor 2 is called factor of 18 because remainder is zero.
18 = 2 ´ 9
2 18
– 18
0
9
Multiple of a given number is the dividend, for
which the given number is its factor.
Thus
The dividend 18 is called multiple of 2 because 2 is factor of 18.

Maths-6
38
Unit 3
Example 1: Is 20 a multiple of 3?
Define even numbers as the numbers which are
multiples of 2
Any number which is multiple of 2, is called an even number.
As 2, 4, 6, 8, 10, 12, ... are multiples of 2.
So, 2, 4, 6, 8, 10, 12, ... are even numbers.
In other words, an even number is a number which is exactly
divisible by 2.
Example 2: Find all multiples of 2.
Example 3: Find the multiples of 3 less than 15.
Example 4: Find the multiples of 4 which are between 9 and 23.
2 ´ 1 = 2
2 ´ 2 = 4
2 ´ 3 = 6
2 ´ 4 = 8
and so on
Solution:
Solution: Multiples of 2 can be found as:
Solution: The multiples of 3 are 3, 6, 9, 12, 15, 18, ...
Solution: The multiples of 4 are 4, 8, 12, 16, 20, 24, 28, ...
So, the multiples of 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, ...
So, the multiples less than 15 are 3, 6, 9, 12
So, the multiples between 9 and 23 are 12, 16, 20.
Since 3 is not a factor of 20
So, 20 is not a multiple of 3.
3 20
– 18
2
6

Maths-6
39
Unit 3
Rule: A number having 0, 2, 4, 6, or 8 at its ones place is an
even number.
Rule: A number having 1, 3, 5, 7 or 9 at its ones place is an
odd number.
(a)
(a)
(a)
(b)
(c)
(d)
odd
even
even
odd
As digit at ones place is 5, which is not even
number. Hence, 3145 is odd number.
As digit at ones place is 4, which is even
number. So, 6784 is an even number.
As digit at ones place is 0, which is even
number. So, 9210 is even number.
As digit at ones place is 1,which is not even
number. So, 2461 is odd number.
(a)
(b)
(c)
(d)
(b)
(b)
(c)
(c)
(d)
(d)
257
3145
257 is not even because digit at ones place is 7.
7208 is even because digit at ones place is 8.
11114 is even because digit at ones place is 4.
15683 is not even because digit at ones place is 3.
7208
6784
11114
9210
15683
2461
Example: Which of the following numbers are even:
Example: Identify the odd or even numbers as the multiple of 2
and explain.
Define odd numbers as the numbers which are not multiples of 2
Solution:
Solution:
A number which is not multiple of 2 is called an odd number.
As 1, 3, 5, 7, 9, 11, 13, ... are not multiples of 2.
So, 1, 3, 5, 7, 9, 11, 13, 15, ... are odd numbers.
In other words
An odd number is a number which is not exactly divisible by 2.
Explanation
3145
6784
9210
2461
Number Type

Maths-6
Define prime numbers as numbers which have only two
factors (i.e., 1 and itself)
Define composite numbers as numbers which have more
than two factors
As 2, 3, 5, 7, 11, 13, 17, 19, ... have only two factors.
So, they are prime numbers.
A natural number which has only two distinct factors 1 and
number itself is called a prime number.
A natural number which has more than two factors is called a
composite number.
As 4, 6, 8, 9, 10, 12, 14, 15, ... have more than two factors.
So, they are composite numbers.
40
Unit 3
(a)
(a) (c)
(a)
(b)
(b)
(b) (d)
23
8 9
23 has two factors 1 and 23. So, 23 is prime number.
8 has four factors 1, 2, 4, 8. So, 8 is not a prime
number.
8
17 5
Example: Which of the following numbers is a prime number?
Example: Identify the composite or prime
numbers with reasons.
Solution:
Solution:
(a)
(b)
(c)
(d)
Composite number
Prime number
Composite number
Prime number
As 1, 2, 4, 8 are the factors of 8,
which are more than two. Hence it is
composite number.
As 1, 17 are the only two distinct
factors so, 17 is prime number.
As 1, 3, 9 are the factors of 9, which
are more than two. Hence it is
composite number.
As 1, 5 are the only two distinct
factors of 5 so, 5 is prime number.
Reasons
8
17
9
5
Number Type

Maths-6
According to the number of factors, there are three types of
natural numbers as given below:
So 2, 3, 5, 7, 11, 13, ... are prime numbers.
If we divide any number by 1, the remainder will always be zero.
as it is clear from the following examples:
So, 1 is factor of every number because it produces 0 as remainder.
(1) 1 is factor of 5
Know that 1 is neither prime nor composite as it has only
one factor which is 1 itself
41
Unit 3
Natural Number
1
2
3
4
5
6
Neither prime nor composite
Prime number
Prime number
Composite number
Prime number
Composite number
Type
1
1, 2
1, 3
1, 2, 4
1, 5
1, 2, 3, 6
Factors
4, 6, 8, 9, 10, 12, 14, 15, ... are composite numbers and 1 is
neither prime nor composite number because it has only one
factor which is 1 itself.
Example: Identify prime, composite and the number which is
neither prime nor composite from the following:
5, 8, 16, 1, 19
Solution:
Prime numbers are 5 and 19
Composite numbers are 8 and 16
The number which is neither prime nor composite is 1
Know that 1 is a factor of every number.

Maths-6
42
Whereas all other even numbers have more than two factors.
We know that all prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, ...
From these prime numbers 2 is only even prime number and all
other are odd.
Unit 3
Hence 2 is the only even number which is also prime.
EXERCISE 3.1
1.
2.
3.
4.
Which of the following are the factors of 20.
Write all the factors of each of the following.
Which of the following are multiples of 6.
Write first seven multiples of:
(a)
(a)
(a)
(a)
(c)
(c)
(c)
(c)
(b)
(b)
(b)
(b)
(d)
(d)
(d)
(d) (e)
(e)
2
15
10
4
5
50
96
12
8
30
18
7
3
125
200
15 20
150
We know that even natural numbers are 2, 4, 6, 8, 10, ....
From these numbers 2 has only two factors which are 1 and 2.
Example: Find factors of 2, 5 and 6 and which number is factor
of each of 2, 5 and 6.
Solution:
Factors of 2 are 1, 2
Factors of 5 are 1, 5
Factors of 6 are 1, 2, 3, 6
Know that 2 is the only even prime number whereas all other
prime numbers are odd
Here ‘1’ is the only number which is factor of each of 2, 5 and 6.

Maths-6
43
Unit 3
7.
5.
6.
8.
9.
10.
11.
12.
13.
Which of the following is even or odd?
Write prime numbers between:
Write an even prime number.
Write all the even numbers between 21 and 51.
Write all the odd numbers between 10 and 40.
Write all the composite numbers less than 40.
Write all the composite numbers between 71 and 101.
Write all the prime numbers between 20 and 60.
Why 1 is neither prime nor composite?
(a)
(a)
(c)
(c)
(b)
(b)
(d) (e) (f)
10
6 and 15
75
60 and 95
47
20 and 40
91 100 117
3.2 TESTS FOR DIVISIBILITY
Test of Divisibility by 2 :
Example: 50, 62, 724, 9126, 3338 are exactly divisible by 2.
Example: 24510 is exactly divisible by 3 because 2 + 4 + 5 + 1 + 0 = 12
and 12 is divisible by 3
A number is exactly divisible by 2 if the digit at units place
of the number is 0, 2, 4, 6 or 8.
A number is exactly divisible by 3 if the sum of the digits of the
number is divisible by 3.
There are rules by which we can check whether a given number
is exactly divisible by 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 15 and 25.
These rules are called tests for divisibility.
Test by inspection whether the numbers 2, 3, 4, 5, 6, 8, 9, 10,
11, 12, 15 and 25 can divide a given number.

Maths-6
44
(Tests for Divisibility)
Unit 3
Example: 213832 is divisible by 8 because 832 is exactly divisible
by 8.
A number is exactly divisible by 8 if the number formed by the
last three digits is exactly divisible by 8 or the last three digits
are zero.
A number is exactly divisible by 9 if the sum of the digits of the
number is exactly divisible by 9.
Example:
9241011 is exactly divisible by 9 because 9 + 2 + 4 + 1 + 0 + 1 + 1 = 18
and 18 is exactly divisible by 9.
Example: 1002316 is divisible by 4 because 16 is divisible by 4.
Example: 5210 and 4115 are exactly divisible by 5.
Example: 2142 is divisible by 6 because it is exactly divisible by
both 2 and 3.
A number is divisible by 5 if digit at the units place of the number
is 0 or 5.
A number is divisible by 6 if the number is exactly divisible by
both 2 and 3.
A number is divisible by 4 if the number formed by the last two
digits of the number is divisible by 4 or the two digits are zeros.

Maths-6
45
Unit 3
Testof Divisibility by 25 :
Example: 26130 is divisible by 15 because it is divisible by both
3 and 5.
Example: 2341625 is divisible by 25 because 25 is divisible by 25.
A number is exactly divisible by 15 if the number is exactly
divisible by both 3 and 5
A number is exactly divisible by 25 if the number formed by last
two digits is exactly divisible by 25 or the last two digits are zeros.
Example: 234084 is exactly divisible by 12 because it is exactly
divisible by 3 and 4 (As 2 + 3 + 4 + 0 + 8 + 4 = 21 is divisible
by 3 and 84 is exactly divisible by 4).
Example:
A number is exactly divisible by 12 if the number is exactly
divisible by both 3 and 4.
A number is exactly divisible by 11 if the difference of the sum of
digits at odd places from the sum of its digits at even places is
either 0 or exactly divisible by 11.
7546 is exactly divisible by 11 because (7 + 4) – (5 + 6) = 11 – 11 = 0
and 907665 is divisible by 11 because (9 + 7 + 6) – (0 + 6 + 5)
= 22 – 11 = 11 which is exactly divisible by 11.
Example: 25670 is exactly divisible by 10.
A number is exactly divisible by 10 if the digit at units place of the
number is zero.

Maths-6
46
Unit 3
Example: Test whether:
Solution:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(i)
(ii)
(iii)
(iv) Here (3 + 5 + 8 + 4) – (6 + 7 + 1 + 6) = 20 – 20 = 0
So, 36578146 is exactly divisible by 11
(v) As 4 + 3 + 2 + 1 + 0 + 2 + 8 + 4 = 24 is exactly
divisible by 3 and 84 is also exactly divisible by 4
So, 43210284 is exactly divisible by 3 and 4
Hence it is exactly divisible by 12
(vi) As 3 + 2 + 4 + 1 + 2 + 0 + 7 + 5 = 24 is exactly
divisible by 3 and there is 5 at ones place
So, 32412075 is exactly divisible by 3 and 5
Hence it is exactly divisible by 15
2412006 is exactly divisible by 3
2 + 4 + 1 + 2 + 0 + 0 + 6 = 15. Since 15 is divisible by 3.
So, 2412006 is divisible by 3.
6123816 is divisible by 8. Since 816 is divisible by 8.
6 + 1 + 7 + 1 + 0 = 15. Since 15 is divisible by 3 and
digit at units place is 0, so it is divisible by 2.
Hence 61710 is divisible by 6.

Maths-6
47
Unit 3
EXERCISE 3.2
1. Test whether.
(i)
(vii)
(iii)
(ix)
(v)
(xi)
(vi) 241566 is exactly divisible by 3
(xii) 1234562 is exactly divisible by 3
(ii) 92348 is exactly divisible by 4
(viii)241361 is exactly divisible by 4
(iv) 24150 is exactly divisible by 8
(x) 4158720 is exactly divisible by 9
2.
3.
5.
6.
4.
Which of the following numbers are exactly divisible by 6?
(i)
(i)
(i)
(i)
(i)
(iii)
(iii)
(iii)
(iii)
(iii)
(ii)
(ii)
(ii)
(ii)
(ii)
2456
24567
215762
245168
25670
19206
111165
94857290
2561742
14675
7121700
230590
52958400
512100
123600

Maths-6
48
Unit 3
7.
8.
9.
10.
11.
Which of the following numbers are exactly divisible
by 2, 3, and 6?
by 4 and 8?
by 2, 5 and 10?
(i)
(i)
(i)
(i)
(i)
(iii)
(iii)
(iii)
(iii)
(iii)
(ii)
(ii)
(ii)
(ii)
(ii)
51300
2563400
14562
513864
120340
236790
4102375
518214
2415848
412360
523449
341236
1101361
617116
210785
3.3 FACTORIZATION
For example: (i) 8 = 2 ´ 4
Here 2 and 4 are
factors of 8
(ii) 12 = 6 ´ 2
Here 6 and 2 are
factors of 12
The process of factorizing a number into its factors is called
factorization.
The process of factorizing a number into its prime factors is
called prime factorization.
Rule: Product of factors of a given number is always equal
to the number.
Define prime factorization as the process of factorizing a
number into its prime factors
Prime factorization can be done by two methods as explained in
the following example.
For example: (i) 8 = 2 ´ 2 ´ 2 (ii) 12 = 2 ´ 2 ´ 3

Unit 3
Example 1: Find the prime factorization of 60:
Solution:
So, 60 = 2 ´ 2 ´ 3 ´ 5
which is prime factorization
of 60
So, 60 = 2 ´ 2 ´ 3 ´ 5
of 60
Division method Tree method
2
2
3
5
60
30
15
5
1
60
´
´
´
´
´
´
2
2
2
2
2
3
15
5
30
Teacher’s Note
Teacher should help the students to practice both methods of prime
factorization.
1
49 Maths-6
Solution:
Example 2: Find the prime factorization of 540:
So,
540 = 2 ´ 2 ´ 3 ´ 3 ´ 3 ´ 5
of 540.
So,
540 = 2 ´ 2 ´ 3 ´ 3 ´ 3 ´ 5
which is prime factorization of 540.
Division method Tree method
2
2
3
3
3
5
540
270
135
45
15
5
1
540
´
´
´
´
´
´
´
´
´
´
´
´
´
´
´
2
2
2
3
3
2
2
3
2
2
2
2
3
3
3 5
270
135
45
15
FACTORS AND MULTIPLES (Factorization)
(Divide 30
by 2)
(Divide
15 by 3)

Maths-6
50
Unit 3
Consider the factors of 360.
360 = 2 ´ 2 ´ 2 ´ 3 ´ 3 ´ 5
So, prime factorization of 360 can be expressed as
The prime factorization of 360 in the above form is known as
index notation.
Here,
and
3
factor 2 is 3 times = 2 (index = 3)
2
factor 3 is 2 times = 3 (index = 2)
1
factor 5 is 1 time = 5 (index = 1)
3 2 1
360 = 2 ´ 3 ´ 5
2
2
2
3
3
5
360
180
90
45
15
5
1
Recognize index notation
Factorize a given number and express its factors in the index
notation
Solution:
Example: Find prime factorization and express it in index notation.
Let us factorize a given number and express its factors in the
index notation with the help of an example.
Hence,
432 = 2 ´ 2 ´ 2 ´ 2 ´ 3 ´ 3 ´ 3
4 3
= 2 ´ 3
Hence,
1500 = 2 ´ 2 ´ 3 ´ 5 ´ 5 ´ 5
2 1 3
= 2 ´ 3 ´ 5
2
2
2
2
3
3
3
432
216
108
54
27
9
3
1
2
2
3
5
5
5
1500
750
375
125
25
5
1
(i)
(i)
(ii)
(ii)
432
432
1500
1500

Maths-6
51
Unit 3
EXERCISE 3.3
2.
1.
3.
Find the prime factorization by division method and
express in index notation.
Write in index notation.
Can you write factorization of a prime number in index
notation.
(i)
(i)
(iii)
(ii)
(v)
(ii)
(vi)
(iii)
(vii)
(iv)
(viii)
24
2 ´ 2 ´ 2 ´ 3 ´ 5 ´ 3
2 ´ 3 ´ 3 ´ 7 ´ 7 ´ 3 ´ 3
3 ´ 5 ´ 5 ´ 5 ´ 7 ´ 7
468
48
540
216
1024
250
5000
3.4 HCF (Highest Common Factor)
Solution:
Hence, HCF = 8
Example: Find HCF of 24 and 40.
HCF of two or more given numbers is the common factor of the
given numbers which is the greatest of all the common factors.
Factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24
Factors of 40 are: 1, 2, 4, 8, 10, 20, 40
Common factors are: 1, 2, 4, 8
The greatest common factor is 8
Define HCF as the greatest number which is a common factor
of two or more numbers.
Find HCF of two or more than two numbers by prime
factorization and long division method
There are two main methods to find HCF. These methods are
explained below:

Maths-6
52
Unit 3
In this method, we use the following rule:
(i) Prime Factorization Method:
HCF = Product of common factors of two or more numbers.
Solution:
Here 48 = 2 ´ 2 ´ 2 ´ 2 ´ 3
36 = 2 ´ 2 ´ 3 ´ 3
40 = 2 ´ 2 ´ 2 ´ 5
Example: Find HCF of 48, 36 and 40.
2
2
2
2
3
2
2
3
3
2
2
2
5
48
24
12
6
3
1
36
18
9
3
1
40
20
10
5
1
Common factors are 2 and 2
So, HCF = 2 ´ 2
= 4
(ii) Long Division Method:
Note: In case of three numbers, we first find HCF of any two
numbers and then continue the process on that HCF and the
remaining number till that last divisor, which is HCF of the all
three numbers.
In case of two numbers, first find the HCF by the method as
given below:
Divide greater number by the smaller number.
Take remainder as divisor and divide the first divisor by it.
Again remainder will be taken as divisor and 2nd divisor
will be treated as dividend.
Continue this process till remainder becomes zero.
The last divisor will be HCF.
FACTORS AND MULTIPLES (HCF)

Maths-6
53
FACTORS AND MULTIPLES (HCF)
Unit 3
Solution:
Solution: First we find HCF of 16 and 36.
Example 1:
Find HCF of 24 and 64 by long division method.
Example 2:
Find HCF of 16, 36 and 70.
So, HCF = 8
So, HCF of 16 and 36 = 4
Hence, the HCF of 16, 36 and 70 is 2.
Now we find HCF of 4 and 70.
24
16
4
64
36
70
– 48
– 32
– 4
24
16
4
8
0
0
HCF
HCF
– 16
– 16
– 4
16
4
30
– 28
2
2
2
17
1
4
2
2
16
0
– 16

Maths-6
54
Unit 3
EXERCISE 3.4
1.
2.
3.
Example: Find the LCM of 3 and 4.
Solution:
Multiples of 3 are: 3, 6, 9, 12 ,15, 18, 21, 24 , 27, 30, 33, 36 , ...
Multiples of 4 are: 4, 8, 12 , 16, 20, 24, 28, 32, 36 , ...
Common multiples of 3 and 4 are: 12, 24, 36, ...
The smallest common multiple = 12
i.e. LCM = 12
Find the HCF of the following numbers by prime
factorization method.
Find HCF of the following numbers by long division
method.
What is the HCF of two prime numbers?
(Take any two prime numbers)
(i)
(i)
(iv)
(iv)
(iii)
(iii)
(ii)
(ii)
(v)
(v)
50, 75
12, 20
120, 144, 204
252, 576
(vii) 60, 70, 420, 480
144, 198
935, 1320
(vi)
(vi)
12, 48, 36, 24
30, 120, 90, 210
98, 196
81, 117
106, 159, 265
2241, 8217, 747
3.5 LCM (Least Common Multiple)
LCM of two or more given numbers is the smallest of all their
common multiples.
Define LCM as the smallest number which is a common
multiple of two or more numbers.

Maths-6
55
FACTORS AND MULTIPLES (LCM)
Unit 3
There are two main methods to find LCM. These are explained
below:
Find LCM of two or more number by prime factorization and
division method
(i)
Note: (i) In case of three numbers, we consider a factor as
common factor if it is common factor of two numbers or of three
numbers.
(ii) Same rule will be applied for more than three numbers.
Solution:
Example: Find LCM of 12, 18, 24.
Here
Here
So, LCM = 2 ´ 2 ´ 3 ´ 2 ´ 3
= 72
Common factors are: 2, 2, 3
Non-common factors are: 2, 3
12 = 2 ´ 2 ´ 3
18 = 2 ´ 3 ´ 3
24 = 2 ´ 2 ´ 2 ´ 3
i.e.
In this method first we find prime factors of each given number
then we find LCM by the following rule.
Prime Factorization Method:
2
2
2
3
2
2
3
2
3
3
24
12
6
3
1
12
6
3
1
18
9
3
1
LCM = Product of common and non-common factors.

Maths-6
56
Unit 3
(ii)
1.
2.
3.
Continue this process until you get “1” as quotient for each
number i.e. (Last row contains all ones)
4.
Then LCM = Product of all divisors.
The main points of this method are given as under.
Division Method:
Choose the smallest divisor that can divide at least one
of given numbers.
Write quotient and the undivided numbers in the second
row.
Again choose the smallest divisor that can divide at least
one of them
Solution:
Example 2:
Find LCM of 12, 15, 50.
Here LCM = 2 ´ 2 ´ 3 ´ 5 ´ 5
= 300
2
2
3
5
5
12, 15, 50
6, 15, 25
3, 15, 25
1, 5, 25
1, 1, 5
1, 1, 1
Solution:
Example 1:
Find LCM of 3, 4, 5.
Here LCM = 2 ´ 2 ´ 3 ´ 5
= 60
2
2
3
5
3, 4, 5
3, 2, 5
3, 1, 5
1, 1, 5
1, 1, 1
Relationship between HCF and LCM
Consider the two number 16 and 24.
Their product 16 ´ 24 = 384
HCF = 8 and LCM = 48
Now multiply HCF and LCM
8 ´ 48 = 384
Thus
The product of any two non zero numbers = The product of HCF and LCM

Maths-6
57
Unit 3
Example: The product of two numbers is 300 and their LCM is
60. What is their HCF?
Solution: The product of two numbers = 300
LCM = 60, HCF = ?
We known that
LCM ´ HCF = Product of two numbers
So, LCM ´ HCF = 300
Thus HCF = 5
60 ´ HCF = 300
HCF =
= 5
300
60
1. Using prime factorization, find the LCM of the following
numbers.
(i)
(iii)
(v)
(ii)
(iv)
(vi)
12, 25 and 40
144, 180 and 384
35, 65 and 75
21, 49 and 63
108, 135 and 162
24, 36, 48 and 72
EXERCISE 3.5
2.
3.
Calculate the LCM of the following numbers by division
method.
The product of two numbers is 360. If their HCF is 6,
what is their LCM?
(i)
(iii)
(ii)
(iv)
45 and 55
24, 40 and 60
21, 35 and 70
72, 108 and 120

Unit 3
The HCF of 520 and 360 is 40
Hence, the required length of the tape is 40 cm.
Here, we have to find HCF
of 60, 80 and 40
63, 85, 47
–3, –5, –7
60, 80, 40
Solution: The greatest length of the tape is the HCF of 520 cm
and 360 cm.
Example 1:
Find the greatest length of a measuring tape which can be used
to measure exactly 520 cm and 360 cm.
Example 2:
Solution:
360 520
– 360
360
40
– 320
160
1
2
4
160
´
– 160
Lets find the required HCF by division method.
Teacher’s Note
Teacher should give more examples related to daily life to students
for their better understanding of applications of HCF and LCM.
1
58 Maths-6
3.6 Applications of HCF and LCM
HCF and LCM are used in daily life as explained in the following
examples.
Solve real life problems related to HCF and LCM
Find the greatest length of wooden scale which can be used to
measure 63 cm, 85 cm and 47 cm leaving remainders 3 cm, 5 cm
and 7 cm respectively.

Maths-6
59
(Application of HCF and LCM)
Unit 3
Solution: The required least quantity of
sugar can be found with the help of LCM.
Solution: First of all we find LCM of 10, 30 and 60.
Here LCM = 2 ´ 2 ´ 2 ´ 3 ´ 5 ´ 5
= 600
Hence the required quantity of sugar is
600 gm.
LCM
So, the required least number of children = 60 + 7 = 67
= 2 ´ 2 ´ 3 ´ 5
= 60
Example 3: Find the least quantity of sugar which can be exactly
measured by 100 gm, 150 gm or 200 gm weights.
Example 4: Find the least number of children which can stand
in rows of 10, 30 and 60 children such that each row has
7 children short.
2
2
2
3
5
5
100, 150, 200
50, 75, 100
25, 75, 50
25, 75, 25
25, 25, 25
5, 5, 5
1, 1, 1
2
2
3
5
10, 30, 60
5, 15, 30
5, 15, 15
5, 5, 5
1, 1, 1
As HCF of 60 and 80 is 20
Now we find HCF of 20 and 40
As the HCF of 60, 80, 40 is 20
So, the required length of
wooden scale is 20 cm.
Lets find the HCF by division method.
60
20
80
40
– 60
– 40
60
0
– 60
20
0
1
2
3

Maths-6
60
Unit 3
EXERCISE 3.6
(Application of HCF and LCM)
1.
2.
Find the greatest length of a wooden scale which can be
used to measure 540 cm and 360 cm exactly.
Find the greatest number which divides 232 and 305
leaving 7 and 5, respectively as remainder.
3.
4.
5.
Find the largest number which divides 245 and 1029
leaving remainder 5 in each case.
Two tankers contain 600 litres and 570 litres of petrol
respectively. Find the maximum capacity of the container
which can measure the petrol of either tanker in exact
number of times.
The length, breadth and height of a room are 8 m, 6 m and
4 m respectively. Determine the longest tape which can
measure the three dimensions of the room exactly.
Solution: First of all we find LCM of 10, 30 and 40.
Here LCM
They will ring together after 120 minutes
together i.e. 120 minutes = 2 hours
Hence the required time = (9 + 2) a.m
= 11 a.m
= 2 ´ 2 ´ 2 ´ 3 ´ 5
= 120
Example 5: Three bells ring at intervals of 10, 30 and 40 minutes
respectively. At what time will they ring together if they start
ringing simultaneously at 9 a.m.
2
2
2
3
5
10, 30, 40
5, 15, 20
5, 15, 10
5, 15, 5
5, 5, 5
1, 1, 1
6. Three bells ring at an interval of 5, 10 and 15 minutes
respectively. If these started ringing at 8 A.M. Find the time
when they will again ring together.

Maths-6
61
Unit 3
8.
9.
Find the least quantity of milk which can be exactly measured
by buckets of capacity 12 litres, 16 litres and 24 litres.
The traffic lights at three different road-crossings change
after every 48 seconds, 72 seconds and 108 seconds
respectively. If they change simultaneously at 7 a.m, after
what time will they change again simultaneously?
1.
2.
3.
4.
5.
6.
7.
8.
9.
7.
Write all the factors of (i) 60 (ii) 250
Write first five multiples of (i) 13 (ii) 20
Write all prime numbers between 1 and 50.
Write all composite numbers between 30 and 60.
Which of the following are divisible by 2:
(i)
(i)
(i)
(i)
(i)
(ii)
(ii)
(ii)
(ii)
(ii)
(iii)
(iii)
(iii)
(iii)
(iii)
(iv)
(iv)
(iv)
(iv)
(iv)
31621
51237
2173
2010
2165
7008
30001
41524
31625
71230
91130
1001001
71611
7128
10000
5178
56712
40048
1001
25618
Find the least number which when divided by 30, 45 and
60 leaves 9 as remainder.
REVIEW EXERCISE 3
10. Find prime factors of the following by division method and
tree method. Also write in index notation.
(i) (ii)
450 720

Maths-6
62
Unit 3
11.
12.
13.
14.
Find HCF and LCM by factorization method:
Find HCF and LCM by division method:
Find the greatest number, that will divide 42, 51 and 67
leaving remainders 2, 1 and 7 respectively.
Find the least number which when divided by 20, 30 and
45 leaving 6 as remainder in each case.
A factor is a number which divides the dividend exactly.
A number has limited number of factors.
Multiple of a given number is the dividend for which the
given number is its factor.
A number has unlimited number of multiples.
A number which is multiple of 2 is called even number.
A number which is not multiple of 2 is called an odd number.
A natural number which has only two distinct factors is
called prime number.
A natural number which has more than two factors is
called composite number.
1 is neither prime nor composite.
1 is factor of every number.
2 is the only even number which is also a prime number.
(i)
(i)
(ii)
(ii)
18, 24
65 and 80
12, 15, 40
26, 65, 169
SUMMARY

Maths-6
63
Unit 3
Divisibility Test a number is divisible by
Product of factors of a given number is always equal to the
number.
HCF = Product of common factors of two or more numbers.
LCM = Product of common and non-common factors.
Product of two non-zero number = LCM ´ HCF
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
(xii)
2 if digit at unit place is 0, 2, 4, 6 or 8.
3 if the sum of digit is multiple of 3.
4 if the number formed by last two digit is divisible by 4.
5 if the digit at unit place is 0 or 5.
6 if the number is divisible by both 2 and 3.
8 if the number formed by last three digits is divisible
by 8.
9 if sum of digits is divisible by 9.
10 if digit at units place is zero.
11 if the difference of sum of digits at odd places from
the sum of digits at even places is either 0 or multiple
of 11.
25 if the number formed by last two digits is divisible
by 25.

Teacher’s Note
Teacher should give some more examples of directed numbers from
daily life.
1
64 Maths-6
4.1 INTEGERS
Know that the natural numbers 1, 2, 3, ..., are also called
positive integers and the corresponding negative numbers
–1, –2, –3, ..., are called negative integers
INTEGERS
Unit
4
5 kilograms of apples,
15 litres of milk etc.
–2 degree centigrade temperature
–5 metres altitude
(5 m below sea level)
Recognize integers
All the integers are as under:
..., –3, –2, –1, 0, 1, 2, 3, 4, ...
Besides positive and negative integers, there is an integer which
is neither positive nor negative that is ‘0’.
Similarly there are corresponding negative numbers –1, –2, –3, .....
which are called negative integers and are used mainly to measure
quantities for example:
We are familiar with natural numbers 1, 2, 3, .... These natural
numbers are also called positive integers and are used mainly to
count and measure the quantities, for example:
The natural numbers 1, 2, 3, ... are also called positive
integers and the corresponding numbers –1, –2, –3, ... are
called negative integers.
‘0’ is an integer which is neither positive nor negative.
Know that ‘0’ is an integer which is neither positive nor negative

By using integers fill in the blanks.
Activity
INTEGERS
A boy and girl starting walking from same point 0. One goes on
right covers 10 m and other goes on left covers 10 m. The starting
point is represented by “0”. The distance of 10 m on the right side
of starting point is shown by +10 m. The distance of 10 m on the
left side of starting point is shown by –10 m.
Example: Fill in the blanks.
–10 m
+8 m
+10 m
(i)
(i)
(ii)
If –10 m represents, distance towards west then __________
represents distance towards east.
If +6 million shows increase in population then __________
shows decrease in population.
(ii)
If +5 m represents, distance of 5 m towards east then
distance of 10 m towards west is ___________.
If –6 m represents distance of 6 m below sea level then the
distance of 8 m above sea level is __________
Unit 4
These integers are also called directed numbers and they are
used to represent distance alongwith the direction or position, as
explained below:
–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10
Left Right
(iii)
(iv)
(v)
If +20 m shows 20 m distance above sea level then _______
shows 50 m below sea level.
If –20 rupees represents loss then _____________represents
profit in a business.
If –50 m represents, distance of 50 m towards south then
___________ represents 60 m towards north.
Maths-6
65

Maths-6
66
INTEGERS
Unit 4
4.2 ORDERING OF INTEGERS
Represent integers on number line
We already know how to represent a natural number on a number line.
Let us represent integers on a number line.
This line is called a number line.
Note: The space between any two points or numbers is always
same.
–3 –2 –1 0 +1 +2 +3 +4 +5
–2 –1 0 +1 +2 +3
–3 –2 –1 0 +1 +2 +3
Know that on the number line any number lying to the right
of zero is positive
Consider a number line.
Let us consider number line.
For example: –1 is to the left of zero. So –1 is negative or –1 < 0.
–2 is to the left of zero so, –2 is negative or –2 < 0 and so on.
On this line we observe that all numbers to the left of zero are
negative.
We observe that all numbers to the right of zero are positive.
For example: +1 is at right of zero or +1 > 0.
+2 is at right of zero which is positive or +2 > 0 and so on.
Thus
Hence
Any number lying to the right of zero is positive.
On a number line, any number to left of zero is negative.
Know that on the number line any number lying to the left
of zero is negative

Maths-6
67
Unit 4
Know that on the number line any number lying to the right
of another number is greater
Let us again consider the number line.
Let us consider the same number line once again.
We observe that any number to the left of any other number is smaller.
For example: –2 is to the left of +1 so, –2 is smaller than +1 i.e. – 2 < +1.
We observe that any number to the right of any other number is
always greater.
For example +3 is to the right of +1. So, +3 is greater than +1.
i.e. +3 > +1
–3 –2 –1 0 +1 +2 +3 +4
Hence
Hence
On a number line, any number to the right of another
number is greater.
On a number line any number to the left of another
number is smaller.
Know that on the number line any number lying to the left
of another number is smaller
–2 –1 0 +1 +2 +3
–6 –5 –4 –3 –2 –1 0 +1 +2 +3 +4 +5
Example: Decide whether first number is greater or smaller using
number line.
Solution:
(i) Let us represent the numbers on number line.
(i) (ii) (iii) (iv)
+5, –2 –6, 0 +4, 0 –3, –5
The number +5 is at the right of –2
Therefore + 5 > –2.
INTEGERS (Ordering of Integers)

Maths-6
68
Unit 4
–6
–6
–5
–5
–4
–4
–3
–3
–2
–2
–1
–1
0
0
+1
+1
+2
+2
+3
+3
+4
+4
+5
+5
+6
+6
(iii) Let us represent the numbers on number line.
(iv) Let us represent the numbers on number line.
The number +4 is at the right of zero.
Therefore +4 > 0.
–6 –5 –4 –3 –2 –1 0 +1 +2 +3 +4 +5 +6
The number –6 is at the left of zero.
Therefore –6 < 0.
The number –3 is at the right of –5.
Therefore –3 > –5.
(ii) Let us represent the numbers on number line.
(i) +5 _________ –5
(ii) +7 _________ +11
(iii) –9 ________ –4
(iv) +2 _______ –20
(<, >)
(<, >)
(<, >)
(<, >)
Fill in the blanks from given options.
Activity 1
Activity 2
(i) –2 is on _________ of +3.
(ii) +5 is on ________ of –10.
(iii) 0 is on _________ of –6
(left, right)
(left, right)
(left, right)
left
>

Maths-6
69
(i) (ii) (iii)
+6 > –7 +1 > –20 +8 > –1
Let us observe the relation between negative and positive
numbers from the following examples:
Let us represent the number on number line.
It is observed that:
From number line it is obvious that all negative numbers are on
the left of positive numbers.
(i) +6 > –7 because +6 is on right of –7
(ii) +1 > – 20 because +1 is on right of –20
(iii) +8 > –1 because +8 is on right of –1
From above examples it is clear that all positive numbers are on
right of negative numbers.
Hence,
Hence,
Unit 4
Every positive integer is greater than a negative integer.
Every negative integer is less than a positive integer.
–20 –16 –10 –8 –7 –5 –1 0 +1 +5 +6 +8 +15
Know that every positive integer is greater than a negative
integer and every negative integer is less than a positive integer
For example: (i)
(ii)
(iii)
(iv)
(v)
–2 < +6
–3 < +1
–36 < +2
+7 > –2
+20 > –50
Arrange a given list of integers in ascending and descending order
On a number line the right most number is the greatest and the
left most number is the smallest.
Given set of integers can be arranged in ascending and descending
order with the help of number line.

Absolute value or numerical value of a number is its distance
from zero on the number line and is always positive or zero.
Maths-6
70
Solution: First of all we indicate the given numbers
( –3, 0, 7, 1, –5) on a number line as under.
Example: Arrange the following numbers –3, 0, 7, 1 –5 in
ascending and descending order.
Here the greatest number is 7 and the smallest number is –5.
So, Ascending order (smallest to greatest) is:
–5, –3, 0, 1, 7
Descending order (greatest to smallest) is:
7, 1, 0, –3, –5
Unit 4
–5 –4 –3 –2 –1 0 +
1 +2 +3 +4 +5 +6 +
7 +8
4.3 ABSOLUTE OR NUMERICAL VALUE OF
AN INTEGER
We know that every integer except zero, on number line
represents its distance from zero alongwith its direction as +5
means distance of 5 units on right of zero and –5 means distance
of 5 units on the left of zero.
If we ignore direction and just consider the distance then this
distance is called absolute or numerical value and we define as:
INTEGERS
Define absolute or numerical value of a number as its distance
from zero on the number line and is always positive
Absolute value of a number is denoted by the symbol “| |”
for example |–6| = 6
we read as “Absolute value of –6 as 6”
Similarly
|+ 7| = 7, |–15| = 15 and |0| = 0

Maths-6
Example:
Write the absolute values of following numbers in ascending and
descending orders.
+4, –2, +1, –3, 0, –5, +6
Absolute values of the given integers are:
So, ascending order is:
And descending order is:
Let us understand the ascending and descending order of
absolute or numerical values of given integers with the help of
example.
Solution:
Here,
4, 2, 1, 3, 0, 5, 6
0, 1, 2, 3, 4, 5, 6
6, 5, 4, 3, 2, 1, 0
|+4| = 4
|–2| = 2
|+1| = 1
|–3| = 3
| 0 | = 0
|–5| = 5
|+6| = 6
and
71
Unit 4
1. Represent the following integers on a number line.
(i) (ii)
(iii) (iv)
–2, –1, 0, +1, +2 –3, –2, –1, 0, +1, +2, +3, +4
–4, –3, –2, –1, 0, +1, +2 +5, –5, –4, +3, +1, –2
EXERCISE 4.1
INTEGERS
(Absolute or Numerical Value of an Integer)
Arrange the absolute or numerical values of the given integers
in ascending and descending order
2. Decide whether first number is greater or smaller.
(i)
(iv)
(ii)
(v)
(iii)
(vi)
+15, –6
–2, –8
–8, 0
+7, +9
+16, 0
–4, –1

Maths-6
72
Unit 4
3.
4.
5.
7.
6.
Arrange the given integers in ascending and descending
order.
Arrange the absolute values of given numbers in
ascending and descending orders.
Write the absolute values of the following.
(i)
(i)
(ii)
(iii)
(iv)
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(i)
(iii)
(ii)
(iv)
(i)
(iii)
(ii)
(iv)
(left, right)
(left, right)
(left, right)
(left, right)
(<, >)
(<, >)
(<, >)
(<, >)
(<, >)
(<, >)
(ii) (iv)
(iii) (v)
–5
–5 is on _______ of + 6
+6 is on _______ of –7
0 is on ________ of –15
0 is on ________ of 20
+10 _______ –20
–16 _______ –4
+25 _______ –100
+30 _______ +50
–17 _______ +17
0 _________ –5
+5, –7, +1, 0, –3, –1
0, –4, +4, –5, +5
–3, +4, 0, –1, +2, +5
–4, –1, –7, –2, –8
–4, +1, –6, +3, 0, +5
–20, –10, +11, +7, 0, –4
–25, 0, +17, –10, +30, –60
+8, –5, +13, –9, –12, +3
+20 –18
0 +50
INTEGERS
(Absolute or Numerical Value of an Integer)
8. Write True or False.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
Positive integer is always greater than negative
integer.
0 is a positive integer.
Every negative integer is less than zero.
Absolute value of –5 is less than absolute of +4.
–5 is smaller than –10.
–25 is greater than –100.
Numerical value of a number can never be
negative.
( )
( )
( )
( )
( )
( )
( )

Maths-6
73
INTEGERS
Unit 4
4.4 ADDITION OF INTEGERS
We have already learnt how to find sum of two or more whole
numbers on the number line.
Let us revise with the help of following examples.
Example 1: Add +2 and +3 using number line.
Explanation:
Starting from zero, first we move 2 steps on right side of zero
reaching at +2.
Here, (+2) + (+3) = +5
From there move 3 steps on right side reaching at +5
Similarly we can add more than two whole numbers as mentioned
in the following example.
Thus, (+2) + (+3) = (+5)
Solution:
–2 –1 0
+2 +3
+5
+1 +2 +3 +4 +5 +6
Example 2: Find the sum: (+4) + (+3) + (+5) using number line.
Here, (+4) + (+3) + (+5) = (+12)
–2 –1 0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +12 +13
+11
+4 +5
+3
+12
Solution:

Maths-6
74
INTEGERS (Addition of Integers)
Unit 4
Example 1: Find the sum of –4 and –3.
Example 2: Find the sum: (–2) + (–5) + (–3).
Here, (–4) + (–3) = –7
Here, (–2) + (–5) + (–3) = –10
Sum of two or more negative integers using number line is
explained by the following examples.
–13 –12 –11 –10 –9 –8 –7 –6 –5 –4 –3 –2
–9 –8 –7 –6 –5 –4 –3 –2 –1 0 +1 +2
–1 +1 +2
0
–3
–3
–4
–7
–10
–5 –2
Explanation: Starting from zero, first we move 4 steps from zero
on its left reaching at –4
Solution:
Solution:
Then from there, we move 3 more steps on left reaching at –7
Thus, (–4) + (–3) = –7
Use number line to display sum of two or more given
negative integers

Maths-6
75
Use number line to display difference of two given positive
integers
The sum of a positive and a negative integer is explained with
the help of following example.
Unit 4
Example: Add +6 and –4.
Here, (+6) + (–4) = +2
First we move 5 steps on right side of 0 reaching at +5.
Then from there, we move 6 steps on right side reaching at +11.
Here, (+5) + (+6) = +11
–2 –1 0 +1 +2 +3 +4 +5 +6 +7 +8
–4
+6
+5
+11
+6
+2
Explanation:
First we move 6 steps from zero on its right reaching at +6.
Sum of two given integers is explained with the help of following
examples.
Example: Add:
(i)
(iii)
(ii)
(iv)
+5 and +6
+8 and –5
–7 and –2
–6 and +2
Use number line to display sum of two given integers
Solution:
Solution: (i) +5 and +6
Then from there, we move 4 steps on left reaching at +2.
Thus, (+6) + (–4) = +2
–2 –1 0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11
_
(4 steps to left = 4)

Maths-6
76
Unit 4
First we move 7 steps from 0 on its left reaching at –7.
Then from there, we move 2 more steps on left reaching at –9.
Here, (–7) + (–2) = –9
–2
–9
–7
Solution: (ii) Add –7 and –2 means to solve (–7) + (–2)
–12 –11 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 +1 +2
First we move 8 steps from zero on its right reaching at (+8).
Then from there we move 5 steps on left; reaching at +3
Here, (+8) + (–5) = +3
Starting from 0, first we move 6 steps from zero on its left,
reaching at –6. Then from there we move 2 steps on its right;
reaching at point –4.
Solution: (iii) +8 and –5
+8
+3
–5
–2 –1 0 +1 +2 +3 +4 +5 +6 +7 +8
Solution: (iv) –6 and +2
–6
–4
+2
–8 –7 –6 –5 –4 –3 –2 –1 0 1 2
+9 +10
Here, (–6) + (+2) = –4

Unit 4
Examples: Solve: (i) (+15) + (–10) (ii) (–20) + (+12)
(i)
Here,
So, (+15) + (–10)
+ (15 – 10)(Sign of integer
wtih greater
absolute
value is +)
+ 5
=
=
|+15| = 15
|–10| = 10
(+15) + (–10) (ii)
Here,
So, (–20) + (+12)
– (20 – 12) (Sign of integer
with greater
absolute
value is –)
=
– 8
=
|–20| = 20
|+12| = 12
(–20) + (+12)
Solution:
Examples: Solve: (i) (+12) + (+13) (ii) (–10) + (–14)
Add two Integers (with unlike signs)
Two integers (with unlike signs) are added using the following
three steps.
(i) (ii)
Here, Here,
Now Now
(+12) + (+13)
+ (12 + 13)
+ 25
(–10) + (–14)
– (10 + 14)
– 24
=
=
=
=
|+12| = 12
|+13| = 13
|–10| = 10
|–14| = 14
(+12) + (+13) (–10) + (–14)
Solution:
(i)
(ii)
(iii)
Take absolute values of given integers.
Subtract the smaller absolute value from the larger.
Give the result the sign of integer with the larger absolute value.
Add two integers (with like signs)
(i)
(ii)
(iii)
Take absolute values of given integers.
Add the absolute values.
Give the result the common sign.
Two integers (with like signs) are added using the following
three steps.
Teacher’s Note
Teacher should ask some easy oral questions for good practice of
addition of integers.
1
77 Maths-6

Maths-6
78
INTEGERS
Unit 4
1.
2.
3.
Find the sum using number line.
Solve the following.
(i)
(iv)
(i)
(iii)
(i)
(iv)
(ii)
(v)
(ii)
(iv)
(iii)
(vi)
(+2) + (+7)
(–8) + (+2)
(+5) + (+2) + (+3)
(+4) + (+5) + (+1)
(+10) + (+20)
(+5) + (–14)
(–5) + (–6)
(+5) + (– 8)
(–5) + (–3) + (–4)
(–2) + (–6) + (–5)
(+7) + (–4)
(+9) + (–9)
(ii)
(v)
(iii)
(vi)
(–15) + (–25)
(–14) + (+8)
(–20) + (–7)
(–30) + (+30)
EXERCISE 4.2
4.5 SUBTRACTION OF INTEGERS
Subtraction is the reverse process of addition because subtracting a
subtrahend from a given number means adding the subtrahend with
reverse direction in given number.
Let us understand with the following example.
So,
Hence in order to subtract +2 from +5 we have added +2 with
reverse direction (i.e. –2) in +5.
So, we say subtraction is the reverse process of addition.
(+5) – (+2)
= (+5) + (–2)
= +3
+5
–2
+3
–2 –1 0 +1 +2 +3 +4 +5 +6 +7 +8
Example 1: Subtract +2 from +5.
Solution: First of all we change the sign of +2 and get –2 then
add –2 in +5 on number line.
Recognize subtraction as the inverse process of addition

Maths-6
79
INTEGERS (Subtraction of Integers)
Unit 4
Example 2: Subtract +3 from +7 using number line.
Example 1: Subtract: (i) +2 from +6 (ii) –3 from +8
Subtract one integer from the other by changing the sign of
the integer being subtracted and adding according to the
rules for addition of integers
It is clear from previous example that if we subtract one integer
from other then we have to follow the following rule.
Here, (+7) – (+3) = +4
Solution: To subtract +3 from +7 means;
to solve: (+7) – (+3)
Solution:
(ii)
Subtract –3 from +8 means to find difference between (+8) and (–3)
Difference
Difference
=
=
=
=
=
=
=
=
(+6) – (+2)
(+6) + (–2)
+ (6 – 2)
+ 4
(+8) – (–3)
(+8) + (+3)
+ (8 + 3)
+ 11
(changing sign and adding)
|+6| = 6
|+8| = 8
|–2| = 2
|+3| = 3
Subtract +2 from +6 means to find difference between (+6) and (+2)
–3 from +8
+7
+4
–3
–2 –1 0 +1 +2 +3 +4 +5 +6 +7 +8 +9
Subtract one integer from the other by changing the sign of
the integer being subtracted and adding according to the rules
for addition of integers.
(i) +2 from +6

Maths-6
80
EXERCISE 4.3
1.
2.
3.
Find the difference using number line.
Subtract.
Solve the following.
Unit 4
Example 2: Solve: (i) (–8) – (–5) (ii) (+7) – (+10)
Solution:
(i)
(ii)
=
=
=
=
=
=
=
=
(–8) – (–5)
(–8) + (+5)
– (8 – 5)
– 3
(+7) – (+10)
(+7) + (–10)
– (10 – 7)
– 3
|–8| = 8
|–10| = 10
|+5| = 5
|+7| = 7
(–8) – (–5)
(+7) – (+10)
(i)
(iv)
(i)
(iv)
(i)
(iv)
(vii)
(ii)
(v)
(iii)
(vi)
(ii)
(v)
(iii)
(vi)
(ii)
(v)
(viii)
(+6) – (+4)
(–9) – (+5)
+6 from –10
15 from +20
(+25) – (–15)
(+40) – (+30)
(–29) – (+17)
(–8) – (–3)
(+9) – (–2)
(–9) – (–3)
(– 7) – (+4)
–10 from –20
–6 from + 14
–8 from –7
–9 from –6
(–30) – (–25)
(–16) – (–18)
(–20) + (–30)
(iii)
(vi)
(ix)
(–11) – (+5)
(+15) – (–20)
(–27) + (+17)
INTEGERS (Subtraction of Integers)
Note: We can solve (+6) + (+2) by adding inversely as under:
(+6) + (+2) = (+6) – (–2) = (+6) + (+2) = +(6+2) = +8

Maths-6
81
4.6 MULTIPLICATION OF INTEGERS
We know that multiplication is the simple form of repeated
addition for example multiplying +3 by +4 means adding +3 four
times i.e. (+3) ´ (+4) = (+3) + (+3) + (+3) + 3
Representation on number line
Recognize that the product of two integers of like signs is a
positive integer
Consider the following examples of multiplication using number line.
(i)
Multiplying +2 with +3 means
adding +2 three times in same
direction of +2.
i.e.
Here (+2) ´ (+3) = +6
From these two examples we observe that.
(–3) ´ (–4) = (+12)
Multiplying –3 with –4 means
adding –3 four times in
opposite direction of –3.
(ii)
(+2) ´ (+3) (–3) ´ (–4)
i.e. (+3) ´ (+4) = +12
INTEGERS
Unit 4
+3 +3 +3 +3
+12
–1 0 +1 +2 +3 +4 +5 +6 +7 +8
+2 +2 +2
–1 0 +1 +2 +3 +4 +5 +6 +7 +8
+3 +3 +3 +3
–1 0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10 +11 +12
+9 +10 +11 +12
Rule 1:
The product of two integers of like signs is a positive integer.
For example:
(i) (ii)
(+3) ´ (+4) = +12 (–3) ´ (–6) = +18
+6 +12

Maths-6
82
Recognize that the product of two integers of unlike signs is
a negative integer
Consider the following examples of multiplication using number line.
(i)
Multiplying –2 with +3 means
adding –2 three times in same
direction of –2.
i.e.
So, (–2) ´ (+3) = –6
From these two examples we observe that.
So, (+4) ´ (–2) = –8
Multiplying +4 with –2 means
adding +4 two times in
opposite direction of +4.
(ii)
(–2) ´ (+3) (+4) ´ (–2)
Unit 4
–2 –2 –2
–7 –6 –5 –4 –3 –2 –1 0 +1
–4 –4
–9 –8 –7 –6 –5 –4 –3 –2 –1 0 +1
Rule 2:
The product of two integers of unlike signs is a negative integer.
In multiplication of two integers we multiply the absolute values
of integers and apply the rules of multiplication of integers.
For example:
(i) (–2) ´ (+5) = –10 (ii) (+3) ´ (–5) = –15
Examples: Find the product.
Solution:
(i) (ii)
(i)
Note: Product of any integer and zero is always zero.
(+15) ´ (–23) (–25) ´ (–14)
(+15) ´ (–23)
– (15 ´ 23) (Applying Rule 2)
– 345
=
=
+ 1 5
´ - 2 3
4 5
3 0 x
- 3 4 5
(ii) (–25) ´ (–14)
+ (25 ´ 14) (Applying Rule 1)
+ 350
=
=
- 2 5
´ - 1 4
1 0 0
2 5 x
+ 3 5 0
INTEGERS (Multiplication of Integers)
–6 –8

Maths-6
83
INTEGERS
Unit 4
4.7 DIVISION OF INTEGERS
We know that the operation of division is the inverse process of
multiplication for example dividing +20 by +5 implies that to find
a number which when multiplied by +5 gives +20 as product.
i.e.
This number is 4.
For example:
(i)
From these two examples we observe that:
So,
(ii)
20 ¸ (+2) = 20 ´ (+ )
20 ¸ (+2) = +5
=
= 5
(–18) ¸ (–3) = (–18) ´ (– )
1
2
1
3
20
2
because
20
20
Same number Guess it?
´ 5
´ 5
=
=
=
=
20
5
20
5
4 4
Recognize that division is the inverse process of multiplication
Recognize that on dividing one integer by another if both the
integers have like signs the quotient is positive
Rule 1: For integers of like signs, the quotient is positive.
For example:
(i) (ii)
(+10) ¸ (+2) = +5 (–20) ¸ (–5) = +4
Remember that
Dividing a number with any other number is in fact the
product of given number with reciprocal of the divisor.
So, (–18) ¸ (–3) = +6
= +
= +6
18
3

Unit 4
Let us consider the following examples of division.
(i)
From these two examples we observe that:
So,
(ii)
(+8) ¸ (–2)
= (+8) ´ (– )
= –
= –4
(+8) ¸ (–2) = –4
(–15) ¸ (+3)
Recognize that on dividing one integer by another if both
the integers have unlike signs the quotient is negative
1
2
8
2
So,
= (–15) ´ (+ )
= –
= –5
(–15) ¸ (+3) = –5
1
3
15
3
Rule 2: For integers of unlike signs, the quotient is negative.
In division of one integer by another, we just perform the division
of their absolute values and apply the rules of division.
For example:
(i) (+6) ¸ (–2) = –3 (ii) (–8) ¸ (+4) = –2
Examples: Find the quotient.
Solution:
(i)
(i)
(–36) ¸ (–4)
(–36) ¸ (–4)
(–36) ¸ (–4)
+ (36 ¸ 4) (Applying Rule 1)
+ 9
=
=
=
(ii) (+54) ¸ (–2)
4 36
– 36
0
9
(ii) (+54) ¸ (–2)
(+54) ¸ (–2)
– (54 ¸ 2) (Applying Rule 2)
– 27
=
=
=
2 54
– 4
–
0
14
14
27
Teacher’s Note
At this level, teacher should only consider division of those integers
which are perfectly divisible.
1
84 Maths-6
INTEGERS (Division of Integers)

Maths-6
85
Unit 4
Know that division of an integer by ‘0’ is not possible
Let us consider division of any integer by 0 for example.
Example: Divide +5 by 0.
Solution:
Dividing +5 by 0 implies to find a number by which when 0 is
multiplied the product should be 5
INTEGERS (Division of Integers)
It is impossible as such number does not exist.
Hence division of any integer by 0 is not possible.
EXERCISE 4.4
1.
2.
3.
Find the product.
Find the quotient.
Write true or false.
(i)
(i)
(i) (ii)
(ii)
(ii)
(iv)
(iv)
(iii) (iv)
(v) (vi)
(vii) (viii)
(v) (vi)
(vi)
(v)
(+15) ´ (–4)
(+25) ¸ (–5)
+5 ¸ 0 = 0 0 ¸ +6 = 0
( ) ( )
(iii)
(iii)
(–16) ´ (–25)
(+100) ¸ (+4)
(+20) ´ (+17)
(–35) ¸ (–7)
(–36) ´ (+12)
(–252) ¸ (+4)
+9 ´ 0 = +9 +6 ´ 0 = 0
( ) ( )
(–10) ¸ (–10) = –1 (–20) ´ (–1) = +20
( ) ( )
(+5) ¸ (–5) = 1 (–18) ¸ (–3) = (+6)
( ) ( )
(+35) ´ (–14) (–43) ´ (–16)
(–126) ¸ (–6)
(–234) ¸ (+3)
0 5
– 0
5
1 0 5
– 0
5
2
, , ____ , ____ , ____
This process of division will continue and never finish.

Maths-6
86
INTEGERS
1.
2.
3.
4.
6.
7.
8.
5.
Write all numbers between –5 and + 5.
Represent the following integers on a number line.
Write the absolute values of:
Solve the following:
Solve the following:
Find the quotient.
Find the difference using number line.
(i)
(i)
(i)
(i)
(i)
(i)
(i)
(iii)
(iii)
(iii)
(iii)
(v)
(v) (vi)
(iii)
(ii)
(ii)
(ii)
(ii)
(ii)
(ii)
(ii)
(iv)
(iv)
(iv)
(iv)
(vi)
–5, –2, 0, +3, +5
–3, +4, –4, +6, 0
(+6) + (–9)
(+6) + (–10)
(+10) ´ (–5)
(+20) ¸ (–4)
(+5) – (+3)
(+9) + (+7) + (–5)
(–8) + (+5)
(+20) ´ (+6)
(–36) ¸ (+6)
(–20) – (–15)
(+6) ´ 0 (–10) ´ (–8)
(+4) – (–3)
–6, –3, –1, +1, +4, +6
+2, –7, +1, 0, +4, –2
(–9) + (–6)
(–10) + (–4)
(–6) ´ (–9)
(–15) ¸ (–3)
(–7) – (–2)
(–1) + (–2) + (3)
(+20) – (–10)
(–15) ´ (+3)
(+40) ¸ (+10)
(+40) – (+25)
REVIEW EXERCISE 4
Unit 4

Maths-6
87
+1, +2, +3, ... are called positive integers.
–1, –2, –3, ... are called negative integers.
0 is neither positive nor negative.
Any number to the right of another number is greater.
Any number to the left of another number is smaller.
Every positive integer is greater than a negative number.
Every negative integer is less than a positive integer.
Absolute value of a number is always non-negative.
Sum of two positive integers is always positive integer.
Sum of two negative integers is always negative integer.
Product of two integers of like sign is always positive integer.
Product of two integers of unlike signs is always negative integer.
Product of zero and any other integer is always zero.
For integers of like signs, the quotient is positive.
For integers of unlike signs, the quotient is negative.
Division by zero is not possible.
Unit 4
SUMMARY
INTEGERS

In regular routine of our life, we add, multiply, subtract or
divide two numbers for different purposes. After each of those
cases we get our result as a simplified number. The process of
getting a simplified number is known as simplification.
SIMPLIFICATIONS
Unit
5
Consider this expression.
We can simplify it by two ways:
5.1 SIMPLIFICATIONS
So to get a unique result of our simplification process, we must
follow some rules of preference; we use brackets to
show the preference of operations to be performed.
3 + 4 ´ 2
If we perform addition of 3 and 4 and then multiply the sum
by 2, we get the answer 14.
If we perform multiplication of 4 and 2 and then add 3 to
the product 8, we get the answer 11.
(i)
(ii)
To know the Kinds of Brackets
(i)
(ii)
(iii)
(iv)
——— Vinculum or bar
( ) parenthesis or curved brackets or round brackets
{ } braces or curly brackets
[ ] square brackets
We know that for simplification brackets are used to group two
or more numbers together with operations.
There are four kinds of brackets.
Maths-6
88

SIMPLIFICATIONS
Maths-6
Unit 5
89
We use following rules to remove the brackets.
(i) If there is a plus ‘+’ sign before brackets, the brackets are
removed without changing the sign of the number within
brackets.
For example, + (2 – 5) = + (–3) = –3
(ii) If there is a minus ‘–’ sign before brackets, the brackets are
removed and the sign of the number within brackets is
changed.
For example, – (2 – 5) = – (–3) = 3
(iii) If there is a number before brackets, the number in the
brackets left after simplification is multiplied by this
number and the brackets are removed.
For example, 4 (2 – 5) = 4 (–3) = –12
Also –5 (3 – 7) = –5 (–4) = + 20
Example 1: Simplify : 5 + {16 – 4 ¸ 2 ´ 3 – (–6 ¸ 2)}
Solution: 5 + {16 – 4 ¸ 2 ´ 3 – (–6 ¸ 2)}
= 5 + {16 – 2 ´ 3 – (–3)}
= 5 + {16 – 6 + 3}
= 5 + (10 + 3)
= 5 + 13 = 18
———
Know the order of preference as, , ( ), { } and [ ], to
remove (simplify) them from an expression.
In problems involving more than one bracket, the brackets
—
should be removed in the order , ( ), { } and [ ]. Removing
brackets means simplifying expression within the brackets to get
the simplified form and then remove the pair of brackets.

SIMPLIFICATIONS
Maths-6
Unit 5
90
5.2 BODMAS RULE
BODMAS indicates the sequence in which more than one
operations are performed in the simplification process of any
mathematical problem.
The chain of letters in BODMAS shows meaning of each
letter as:
B
O
D
M
A
S
for
for
for
for
for
for
“Brackets”
“order of operation”
“Division”
“Multiplication”
“Addition”
“Subtraction”
Removal of brackets and solving of operation ‘of’ with operations
of DMAS is known as BODMAS rule.
Example 2: Simplify : 2 [–2 {4 ´ 9 ¸ 8 (5 – 3 – 4)}]
Solution: 2 [–2 {4 ´ 9 ¸ 8 (5 – 3 – 4)}]
= 2 [–2 {4 ´ 9 ¸ 8 (5 – (–1)}]
= 2 [–2 {4 ´ 9 ¸ 8 (5 + 1)}]
= 2 [–2 {4 ´ 9 ¸ 8 ´ 6}]
= 2 [–2 [4 ´ ´ 6]
= 2 [–2 { }]
= 2 [–2 {9 ´ 3}]
= 2 [–2 ´ 27]
= 2 ´ –54
= –108
9
8
4 ´ 9 ´ 6
8
3
1
2
Recognize BODMAS rule to follow the order in which the
operations, to simplify mathematical expressions, are
performed.

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